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-1 




























Basic Engineering Thermodynamics 



BASIC ENGINEERING 
THERMODYNAMICS 



Vincent W. Young 

Consultant in Thermodynamics 


Formerly Professor of Mechanical Engineering 
Oklahoma Agricultural and Mechanical College 


FIRST EDITION 


New York Toronto London 
McGRAW-HILL BOOK COMPANY, INC. 

1952 



BASIC ENGINEERING THERMODYNAMICS 


Copyright, 1952, by the McGraw-Hill Book Company, Inc. Printed in the 
United States of America. All rights reserved. This book, or parts thereof, 
may not be reproduced in any form without permission of the publishers. 

Library of Congress Catalog Card Number : 51-12658 


COPy ^gst 




THE MAPLE PRESS rOMPA MV j YORK, PA. 



PREFACE 


Thermodynamics is prerequisite to the specialized fields of engineering 
study. It may be designated as the key subject in the field of mechanical 
engineering, unlocking the door to more advanced courses. Because of its 
importance, many books are written on the subject of engineering thermo¬ 
dynamics. If these books are to be suitable for use as texts by the under¬ 
graduate engineering student, much of their content must consist of a 
rather conventional review of the principles of thermodynamics; the 
author can, at best, expect to contribute only occasional ideas and meth¬ 
ods that are new. 

The value of a new text in engineering thermodynamics should therefore 
be estimated with regard to factors other than novelty of treatment. 
The author of this text feels that the items in terms of which an appraisal 
should be made would include the following, not necessarily in the order of 
their importance. 

1. Selection of material . Only an introduction to a subject as extensive 
as engineering thermodynamics can be accomplished in a book of a practi¬ 
cal length for use as a text. Yet a foundation must be built for all of the 
specialized courses which are to follow, and a careful selection must be 
made to avoid an uneconomical use of the student’s time. 

2. Continuity of treatment. A selection of material having been made, 
the next requirement is that this material shall be so organized as to result 
in a continuous development of the main theme. The temptation to con¬ 
tinue the development at hand beyond the point which is appropriate to 
the immediate purpose must be resisted in order to avoid distractions from 
the basic argument. 

3. Balance and consistency. There are few authors who do not have 
their specialties within the broad field of engineering thermodynamics, 
and there is always a temptation to discuss these fields of special interest 
at greater length than is appropriate to the over-all purpose of the text. 
On the other hand, a tendency toward too brief and condensed a survey of 
topics of less interest to the writer must also be avoided. 

4. Rigor of treatment. The development of thermodynamic concepts 
should be sufficiently rigorous to satisfy the student of the validity of the 
reasoning and convince him of the soundness of the argument. An under¬ 
graduate text may easily follow a line that is too rigorous, however, 
defeating its own purpose by presenting a proof that the average under¬ 
graduate student follows with difficulty, if at all. 


VI 


PREFACE 


5. Readability. The engineering student is inclined to resent a text 
which employs a style that, no matter how precise and rigorous are its 
arguments, requires that he spend too much time in rearranging and 
rephrasing its sentences and paragraphs into a form in which he can grasp 
their meanings. He is right, of course, for in analyzing the language used 
and in translating it into the form that makes sense to him, he may 
be distracted from the continuity of thought in which he should be 
encouraged. 

6. Clarity. A simple and uncomplicated style of presentation is not the 
only requirement for clarity. In addition, the discussion of a given topic 
should be neither too condensed nor too extended. In this respect, the 
teacher of thermodynamics has a distinct advantage in writing on the sub¬ 
ject. Through the years, the questions which his students have asked 
have told him what topics require more detailed and forceful presentation 
than would appear necessary on the basis of superficial examination, and 
he is thereby guided in the emphasis which he places upon these items. 
They will also have shown him when a numerical example will comple¬ 
ment a general discussion of the topic to good effect. 

The author of this text has had the objectives outlined above constantly 
in mind during the period of its compilation. He can only hope that those 
who read his book will feel that he has, in the main, attained them. The 
book has been designed as the basis of a two-semester course in engineer¬ 
ing thermodynamics, a total of six to eight semester hours. By less inten¬ 
sive coverage and by eliminating certain parts, it may be used for a one- 
semester course. The treatment is such that the text is suitable for indi¬ 
vidual study, and the chapters not covered in classwork may be used for 
reference during later specialized courses and more advanced study. The 
author will welcome suggestions for the improvement of the book and 
notification of any errors that may be found. 

In a few instances the method of approach differs from the usual. For 
example, the early identification of the ideal process as a maximum-work 
process and the use of the maximum-work principle as a tool throughout 
the book are a variation from the usual practice. The author has used 
this method in his classes and has found that it is a concept that makes 
sense to the student and is effective with him. 

A full acknowledgment of credit is always impossible to the author of a 
book of this character. He has been influenced by his teachers, by the 
books he has read, by his students, by the texts from which he has taught. 
Credit is given specifically for the use of certain references at the points 
where the material has been introduced into this text. There are other 
places where the line of thought has been appropriated unconsciously and 
applied to the author’s purpose. Special thanks are due Professor James 
H. Boggs of the faculty of the Oklahoma Agricultural and Mechanical 


PREFACE 


Vll 


College for his assistance in reading and criticizing the manuscript, to 
Professor Bess Allen, Oklahoma Agricultural and Mechanical College, for 
her advice on matters of phrasing and English, and to the author’s wife, 
Katherine, without whose understanding cooperation the writing of this 
book would have been a difficult task indeed. The author also wishes to 
thank Professor Emeritus Edwin F. Church, Jr. of the Polytechnic Insti¬ 
tute of Brooklyn, whose especially able and constructive criticism of the 
manuscript has made it possible to correct and improve this text prior 
to its publication. 

Vincent W. Young 

Whittier, Calif. 

January, 1952 






































































CONTENTS 


Preface . v 

1. FUNDAMENTAL DEFINITIONS AND CONCEPTS. 1 

!• Thermodynamics. 2. The thermodynamic system. 3. Stored energy. 

4. Properties of the system. 5. The pure substance and the simple system. 

6. The state path. 7. The process. 8. The cycle. 9. Identification of a 
property. 10. The equation of state. Problems. Symbols. 

2. THE FIRST LAW AND THE CLOSED SYSTEM.21 

1. The First Law of thermodynamics. 2. The closed-system process. 3. 

Work and the closed system. 4. Maximum work on a piston. 5. Useful 
work. 6. Constant-volume processes. 7. The specific heat at constant 
volume. 8. Constant-pressure processes. 9. The specific heat at con¬ 
stant pressure. 10. The constant-temperature (isothermal) process. 11. 
Constant-internal-energy processes. 12. The adiabatic process. 13. The 
paddle-wheel process and friction. 14. The closed-system cycle. 15. The 
heat engine. 16. Efficiency of a heat engine. Problems. Symbols. 

3. THE FIRST LAW AND THE OPEN SYSTEM.44 

1. Stored energy of the open system. 2. Work and the open system. 3. 

Flow work. 4. The First Law and the open system. 5. Steady flow. 6. 

The Bernoulli equation. 7. The continuity equation of steady flow. 8. 

The steady-flow heat engine. 9. Quasi-steady flow. Problems. Symbols. 

4. THE REVERSIBLE PROCESS AND THE REVERSIBLE CYCLE . . 60 

1. Thermodynamic reversibility. 2. Reversibility and the closed-system 
process. 3. The reversible adiabatic closed-system process. 4. The rever¬ 
sible isothermal process. 5. Reversible heat flow at variable temperature. 

6. The constant-volume reversible process. 7. External and internal 
reversibility. 8. Reversibility in steady flow. 9. The Carnot engine. 

10. The Stirling cycle and Stirling engine. Problems. Symbols. 

5. THE SECOND LAW OF THERMODYNAMICS.78 

1. Inadequacy of the First Law. 2. The Second Law. 3. Reversibility 

and the Second Law. 4. The Second Law and the Carnot engine. 5. The 
Carnot principle. 6. The thermodynamic scale of temperature. Problems. 
Symbols. 

6. ENTROPY, A PROPERTY OF THE SYSTEM.88 

1. Entropy. 2. The temperature-entropy diagram and the reversible proc¬ 
ess. 3. The temperature-entropy diagram and the irreversible process. 4. 

The enthalpy-entropy (Mollier) chart and the steady-flow process. 5. The 
temperature-entropy diagram and the reversible cycle. 6. Entropy as a 
criterion of stability. 7. Internal irreversibility. 8. The availability and 
unavailability of heat. 9. The psi function. 10. The zeta property. 
Problems. Symbols. 


IX 




X 


CONTENTS 


7. THE PURE SUBSTANCE.117 

1. Phases of the pure substance. 2. The pvT surface. 3. The phase dia¬ 
gram. 4. The pressure-volume diagram. 5. The triple point. 6. The 
critical point. 7. Tables of properties. 8. The temperature-entropy dia¬ 
gram. 9. The Mollier diagram. 10. The experimental determination of 
quality. Problems. Symbols. 

8. GENERAL THERMODYNAMIC EQUATIONS FOR THE PURE SUB¬ 
STANCE .150 

1. The Maxwell relations. 2. The specific heats. 3. The general equa¬ 
tions for enthalpy, internal energy, and entropy. 4. The Clapeyron rela¬ 
tion. Problems. Symbols. 

9. THE PERFECT GAS.167 

1. The gas and the vapor. 2. The perfect gas. 3. Joule’s law. 4. The 
ratio of the specific heats. 5. Reversible nonflow processes. 6. The irre¬ 
versible nonflow process. 7. Changes of entropy. 8. The steady-flow 
process. 9. The continuous compression of gases. 10. Multistage com¬ 
pression. 11. Clearance factor and volumetric efficiency. 12. Steady-flow 
expansion of gases. Problems. Symbols. 

10. MIXTURES OF GASES AND VAPORS.203 

1. Basic principles. 2. Definitions. 3. Calculation of mixture properties. 

4. Adiabatic mixing processes for perfect gases. 5. The isentropic process. 

6. Characteristics of gas-vapor mixtures. 7. Definitions. 8. The psy- 
chrometric chart. 9. Engineering processes for atmospheric air. Prob¬ 
lems. Symbols. 

11. STEADY FLOW OF FLUIDS—THE TURBINE.229 

1. Introduction. 2. The reversible adiabatic nozzle. 3. The liquid nozzle. 

4. The perfect-gas nozzle. 5. The vapor nozzle. 6. The real nozzle. 7. 
Pressure variation in the convergent-divergent nozzle. 8. Metastable 
expansion—supersaturated steam. 9. Turbine staging. 10. Impulse stag¬ 
ing. 11. Reaction staging. 12. Comparison of impulse and reaction stag¬ 
ing. 13. Stage efficiency. 14. Turbine efficiency. 15. The axial-flow 
compressor. Problems. Symbols. 

12. POWER—GAS SYSTEMS.268 

1. Introduction. 2. The air-standard cycle. 3. The Otto engine. 4. The 
diesel cycle. 5. The dual cycle. 6. Comparison of Otto-, diesel-, and 
dual-cycle efficiencies. 7. The Lenoir engine. 8. The Brayton (gas-tur¬ 
bine) cycle. 9. Improvements in the gas-turbine power-plant cycle. 10. 

Jet propulsion. 11. The Holzwarth cycle and the explosion gas turbine. 

12. External combustion and the closed cycle. Problems. Symbols. 

13. THE RECIPROCATING STEAM ENGINE.305 

1. The reciprocating engine. 2. The reversible adiabatic engine. 3. The 
real-engine cycle. 4. Steady-flow analysis of the reciprocating engine. 5. 
Engine efficiency. 6. Sources of irreversibility in the reciprocating engine. 

7. Hirn’s analysis. 8. Reduction of cylinder condensation. 9. Methods 
of governing the engine. 10. Reciprocating engine versus turbine. Prob¬ 
lems. Symbols. 

14. POWER—VAPOR SYSTEMS.327 

1. The Carnot cycle as a vapor cycle. 2. The Rankine cycle. 3. The 
reheat cycle. 4. The regenerative vapor cycle. 5. The ideal vapor for the 










CONTENTS 


xi 


heat engine. 6. The mercury-steam binary-vapor cycle. Problems. 
Symbols. 

15. REFRIGERATION.358 

1. Introduction. 2. The reversed Carnot as a refrigeration cycle. 3. The 
vapor-compression cycle. 4. Refrigerants. 5. Absorption refrigeration. 

6. Adsorption refrigeration. 7. Regeneration in refrigeration-liquefying 
gases. 8. The heat pump. Problems. Symbols. 

16. THE EVALUATION OF IRREVERSIBILITY.386 

1. Introduction. 2. Availability. 3. The evaluation of irreversibility. 4. 
Availability in steady flow. 5. Irreversibility in steady flow. 6. Efficiency 
versus effectiveness of the turbine stage. Problems. Symbols. 

17. THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS. ... 408 

1. Introduction. 2. Viscosity and viscous drag. 3. Dimensional analysis. 

4. Effect of Reynolds number on flow characteristics. 5. Pressure drop in 
pipes and ducts. 6. Fanno lines. 7. The Rayleigh line. Problems. 
Symbols. 

18. REAL GASES.433 

1. Introduction. 2. The perfect gas with variable specific heat. 3. The 

gas tables. 4. The van der Waals equation. 5. Other general equations of 
state. 6. The law of corresponding states. 7. The generalized Z chart. 8. 

The properties of a real-gas mixture. 9. Conclusion. Problems. Symbols. 

19. THERMODYNAMICS OF COMBUSTION.458 

1. Introduction. 2. The combustion equation. 3. Analysis of products. 

4. The heat of combustion. 5. The heat of formation. 6. Dissociation. 

7. The flame temperature. 8. Power from combustion. Problems. 
Symbols. 

20. THE TRANSMISSION OF HEAT.487 

1. Introduction. 2. Heat transfer by conduction. 3. Thermal conductiv¬ 
ity. 4. Typical problems in heat transfer by conduction. 5. Heat transfer 

by radiation. 6. The Stefan-Boltzmann equation. 7. Radiant-heat trans¬ 
mission of gases. 8. Effect of convection on the transmission of heat by 
conduction—film coefficients. 9. The over-all coefficient of heat transmis¬ 
sion, U. 10. The application of dimensional analysis. 11. Calculation of 
film coefficients—laminar flow. 12. Mean temperature difference. Prob¬ 


lems. Symbols. 

Appendix .529 

Index .531 








































































































CHAPTER 1 

FUNDAMENTAL DEFINITIONS AND CONCEPTS 


1:1. Thermodynamics is the science which is concerned with changes 
in the form or the location of energy. The subject is of special importance 
to the engineer since he deals with energy in large quantities. 

The structure of thermodynamic theory is founded on the law of con¬ 
servation of energy, which affirms that energy can be neither created nor 
destroyed and that structure would at once collapse if this law were shown 
to be invalid. Although the proof of the law is of a negative rather than 
a positive character (an instance has never been observed in which it did 
not hold), it is extremely unlikely that this foundation is not sound and 
solid; it has been time-tried and tested. 

In general, the procedure in thermodynamic analysis consists in segre¬ 
gating a body of matter, called the thermodynamic system , which has the 
ability to receive, to store, and to release energy and in tracing it through 
the thermodynamic process that results as it changes its state. Attention 
is especially directed toward the amount and form of the energy that 
enters or leaves the system and to changes in the amount of energy stored 
in the system during the process. By the law of conservation of energy, 
the change in the stored energy of the system must derive directly from 
the energy which has been transferred across its boundaries. 

1:2. The thermodynamic system is composed of a limited portion of 
matter and is defined in terms of the boundaries which enclose it. The 
two types of systems that are employed in thermodynamic theory are the 
closed system and the open system. 

The boundaries of a closed system permit the passage of energy, but no 
matter may cross them. Thus the mass contained within these bounda¬ 
ries is constant 1 though the boundaries themselves may change their posi¬ 
tion to allow for changes in the volume or the shape of the system during 
a prescribed change of state. The state or condition of the closed system 
at a given instant is described in terms of its properties. 

1 Note that the weight is not constant if the gravitational constant g changes. It 
will be assumed throughout this text that the acceleration due to gravity is constant 
and that unit mass and unit weight may therefore be identical. The unit of mass we 
shall usually employ will be the pound, an amount of mass that, as the result of the 
earth’s attraction, would weigh 1 pound; mass in terms of this unit will be denoted by 
the symbol M. Another unit of mass is the slug, which, under the standard accelera¬ 
tion of gravity, weighs 32.17 pounds. In terms of this unit, mass will be denoted by 
m; thus M = m/g = m/32.17. 


1 


2 


BASIC ENGINEERING THERMODYNAMICS 


Energy that crosses the boundaries of a closed system will be classified 
as either heat or work. Heat is the form of energy which is transferred 
from external systems at higher temperature by reason of the temperature 
difference. The standard convention assigns the 'positive sign to heat 
that enters the system (owing to the system being at lower temperature 
than external systems) and the negative sign to heat leaving the system. 

In thermodynamic usage, the term heat is restricted in its application 
to energy crossing the boundaries of the system solely by reason of a tem¬ 
perature difference between the system and external systems with which 
it is in communication. It is obvious that the state of the system might 
be so described as to prescribe a temperature difference between various 
parts of the system within its assigned boundaries. In that case the sys¬ 
tem is not in a state of equilibrium , and a tendency will exist toward a 
spontaneous exchange of energy within the boundaries due to this internal 
temperature difference. In the sense in which the term is used in thermo¬ 
dynamics, this exchange between parts of the same system does not con¬ 
stitute a flow of heat; it is rather to be considered a redistribution of 
stored energy. It is, of course, hardly necessary to state that stored 
energy in any form cannot properly be called heat. Heat is a transitory 
form of energy; energy cannot exist as heat. When heat leaves the sys¬ 
tem, we have, as evidence of its passage, an immediate increase in the 
stored energy of external systems; we may observe heat only in terms of the 
effects that its transfer produces. Heat is measured in quantity by its 
effects in terms of changes of stored energy as, for instance, by the change 
of temperature which it can produce in a given mass of a prescribed 
material. 

The term work will undoubtedly recall to the mind of the reader the 
meaning of the word as used in mechanics. In mechanics, work was 
defined as that form of energy which resulted when a force moved through 
a distance and was measured as the product of the force and the distance 
it moved in its own direction. Thermodynamically, the term has both a 
broader and a narrower significance. Its meaning is broader in thermo¬ 
dynamics because it would include any form of transitory energy that 
might have as its only effect the type of work which we associate with 
mechanics. Thus the flow of electricity would be classified as work in the 
thermodynamic sense for we know from a study of electrical energy that 
it is, theoretically and in the limit, completely and continuously converti¬ 
ble to the work of mechanics. 

On the other hand, the meaning of the term work is narrower in the 
thermodynamic sense, for, to be classified as work, it must cross the bound¬ 
aries of the system in that form. This means that if a system expands it 
does work only if it meets and overcomes resistance from external sys¬ 
tems; the amount of work performed is measured as the product of that 


FUNDAMENTAL DEFINITIONS AND CONCEPTS 


3 


resistance and the distance (in its own direction) through which it is over¬ 
come. If a system is not in equilibrium, the pressure of its various parts 
may not be uniform and that part at higher pressure may expand, com¬ 
pressing the rest of the system; note that this is not work in the thermo¬ 
dynamic sense, for this energy does not cross the system boundaries. 
The standard convention gives the positive sign to work delivered by the 
body to external systems, the negative sign to work received by the system 
as, for example, when it is compressed. 

An open system has boundaries which permit the passage not only of 
heat and work but also of matter. Thus the mass contained within those 
boundaries may change during a process. Also, the total change in the 
stored energy of an open system will depend not only upon the energy in 
the form of heat and work which has crossed its boundaries during the 
change of state but also upon the energy that crossed those boundaries in 
the form of stored energy in the matter which was introduced into or 
exhausted from the system during the same interval. However, this 
stored energy, though it crosses the system boundary, is not to be classi¬ 
fied as transitory energy; heat and work are the only forms of transitory 
energy in the sense in which that word is used in thermodynamics. In 
the more common applications of the open system, new material is 
introduced at one or more points on the boundaries of the open sys¬ 
tem at exactly the same rate as it leaves at another point or points. 
In this special case the mass included within the system boundaries 
does not change although it is not composed of the same particles of 
matter. 

The study of the thermodynamic process requires that the boundaries 
of the system must be accurately known. In later problems in this book, 
the extent of the system will often be defined by italicizing the words 
which describe its content. 

1:3. Stored Energy. In order to understand the various forms in 
which energy may be stored in a system, it is necessary to set up some sort 
of concept as to the structure of the matter of which the system is com¬ 
posed. The kinetic theory of matter will be used for that purpose. But 
even if this theory, generally accepted as it is, were replaced by another, 
the structure of thermodynamic theory would not be endangered as long 
as the possibility of energy storage in matter, as implied in the law of 
conservation of energy, was not denied. 

Matter is defined as that which has mass and occupies space. It may 
exist in the solid, the liquid, or the gaseous phases. As a solid the system 
has at least a certain amount of rigidity, a definite volume, and a dis¬ 
tinctive shape which can be changed only by the application of apprecia¬ 
ble force. The liquid assumes the shape of the vessel that holds it but 
retains a definite volume, being bounded by a free surface; it is compressi- 


4 


BASIC ENGINEERING THERMODYNAMICS 


ble in only slight degree. A gas has neither shape nor volume of its own 
but fills any container in which it may be held; it is readily compressible. 

According to the kinetic theory, the system is composed of very small 
particles called molecules. The number of these molecules is tremendous; 
for example, there are about 5 X 10 20 molecules in a cubic inch of atmos¬ 
pheric air. But, in spite of their very large numbers, the individual mole¬ 
cule is so small that, even in the case of the solid, there are relatively large 
open spaces between them. Each molecule is attracted to every other 
molecule by a force which is analogous to the attraction between the 
planets of the solar system and which is inversely proportional to the 
square of their distance apart. It would seem that these attractive 
forces would ultimately draw the molecules of a system into a compact 
mass, but the individual molecule also has motion. This motion may be 
a translation or a rotation. In translation, the molecule travels through 
the voids between other molecules until it collides with another molecule, 
after which each travels off in a new direction at undiminished speed. 
Their speeds will vary with the weight of the molecule and with the tem¬ 
perature of the system and range upward to over a mile a second for 
hydrogen gas at ordinary atmospheric temperatures. The speed decreases 
with greater molecular weight and reduced temperature. 

The molecule, in turn, is composed of atoms. A few molecules, the 
argon and helium molecules, for example, consist of only a single atom. 
Thus argon and helium are called monatomic gases. On the other hand, 
the number of atoms per molecule may be quite large. These atoms con¬ 
tain a heavy central nucleus with a positive charge of electricity about 
which negatively charged and extremely light electrons revolve. 

The thermodynamic system as a whole may store energy in two forms; 
these are called potential energy and kinetic energy. Potential energy of 
the system is exemplified by the type of energy possessed by the water in 
an elevated tank due to its height above some datum level and is measured 
as the amount of work required to raise it to that elevation from the datum 
plane. Another example is the energy possessed by an extended spring 
due to the stresses set up in its material as it is lengthened; again the 
amount of potential energy of the spring is relative to some reference 
condition and is measured by the work the spring has received in extend¬ 
ing it from the reference condition. The systems ordinarily encountered 
in engineering thermodynamics are of low density, and differences of ele¬ 
vation are so minor that the potential energy of the system may often be 
ignored with negligible error. 

A system that is in motion possesses kinetic energy whether the motion 
is translational or rotational. Simple examples are a thrown ball, a flow¬ 
ing fluid, or a rotating flywheel. The second of these examples would be 
of more interest in engineering thermodynamics since the systems 


FUNDAMENTAL DEFINITIONS AND CONCEPTS 


5 


employed are usually composed of liquids or gases and because instances 
of translational movement of the system as a whole are much more com¬ 
mon than rotation. The translational kinetic energy of a system is a 
function of its mass and the square of its velocity. Velocity must be 
expressed with respect to some other system such as the surface of the 
earth, and the kinetic energy of the system is thus, like potential energy, a 
relative quantity. The kinetic energy of the system as a whole is readily 
and completely convertible to or from work, and its amount may be 
measured in terms of the work required to impart the given velocity to 
the system. 

It was emphasized in the two paragraphs immediately above that we 
were discussing the stored energy of the system as a whole and that this 
form of stored energy was completely convertible into work. For pur¬ 
poses of easy reference, we shall hereafter often speak of the stored energy 
of the system as a whole, whether potential or kinetic, as stored mechanical 
energy. 

Stored mechanical energy does not take any account of the energy 
stored in the molecule because of its motion or its position with respect to 
other molecules. When the average distance between the molecules of a 
closed system is increased owing to an expansion of the boundaries of the 
system, an increase in the energy storage of this collection of molecules is 
indicated. This is analogous to the type of energy stored in a spring as it 
is extended (potential energy) but differs from it in an important respect. 
The pull between molecules is not concentrated in a single direction; each 
molecule has an attraction for other molecules on all sides of it. It might 
be considered an unorganized form of potential energy and certainly is one 
which is not readily convertible into work. We measure it ordinarily in 
terms of its thermal effects. In the case of gases, especially at low pres¬ 
sure and correspondingly large volume, the distance between molecules is 
relatively very much larger and the attractive forces between them corre¬ 
spondingly smaller. Accordingly, when a gas system changes its volume 
(but remains in the gaseous phase), there is very little change in stored 
potential molecular energy; in fact it may be possible to ignore changes in 
this form of stored energy altogether in dealing with this class of system. 
Of course this is not possible when the system changes its phase during the 
thermodynamic operation. 

The velocity of the molecule, whether translational or rotational, repre¬ 
sents stored energy analogous to the kinetic mechanical energy of a 
thrown ball, rotating in flight. But again this is an unorganized form of 
stored energy since the molecules are traveling in all directions. It is 
neither readily nor completely and continuously convertible to work and 
is measured in terms of its thermal effects, e.g., in terms of the heat flow 
across the boundaries of the system which accompanies a change in the 


6 


BASIC ENGINEERING THERMODYNAMICS 


amount of this form of stored energy. Again for purposes of ready refer¬ 
ence, we shall speak of the stored energy of the molecule, whether poten¬ 
tial or kinetic in character, as stored thermal energy. In the thermo¬ 
dynamic transaction, accent is placed on the changes of stored thermal 
energy that take place in the system rather than the absolute amount ol 
energy stored in this form. 

The engineer is often concerned with combustion. This is a type of 
chemical reaction during which, as the result of a new alignment of atoms 
and, at times, a change in the total number of molecules that constitute 
the system, energy is released. As an example, consider the union of two 
molecules of hydrogen, each composed of two hydrogen atoms (2H 2 ), 
with one oxygen molecule, consisting of two atoms of oxygen (0 2 ), to 
form two molecules of water, each comprising two atoms of hydrogen and 
one of oxygen (2H 2 0). The chemist would symbolize this reaction in the 
form 

2H 2 + 0 2 —» 2H 2 0 + Q 

in which Q represents what he calls the “heat of reaction.” Recalling 
that in thermodynamics the term heat can never properly be used except 
as energy crosses the boundaries of a system, we observe that the expres¬ 
sion heat of reaction must be used with care and understanding in engi¬ 
neering thermodynamics. To illustrate, let us suppose that the reaction 
which has been used as an example took place within a closed, rigid, non¬ 
conducting envelope. Although the mass of the matter contained within 
that envelope has not changed as the result of the reaction (the number of 
each kind of atom is the same in the products as in the reactants), there 
has been a reduction of one-third in the number of molecules since each 
molecule of the products contains three atoms instead of the two per mole¬ 
cule of each of the reactants. Because the envelope is rigid and noncon¬ 
ducting, no energy may pass the boundaries of the closed system con¬ 
tained within it. In the cited example, a small portion of the so-called 
“heat of reaction” may be absorbed owing to increasing the average dis¬ 
tance between molecules. Even this is not always the case, for in some 
combustion reactions the number of molecules in the products may equal 
or even exceed the number in the reactants. Most of this energy must 
be taken up by an increased kinetic energy of the molecules, which means 
greater rates of molecular movement. It has already been noted that a 
higher temperature accompanies greater molecular velocities, and we may 
accordingly expect an increase of temperature. If now the envelope is 
made conducting (though still rigid), heat will pass outward across the 
boundary of the system because the temperature of the system is higher 
than that of its environment. Assuming that the environment is vast and 
thus capable of receiving this heat without change of temperature and 


FUNDAMENTAL DEFINITIONS AND CONCEPTS 


7 


that the constant temperature of the environment is the same as the 
original temperature of the reactants, this passage of heat will continue 
until the temperature of the products is the same as the original tempera¬ 
ture of the reactants. The total heat flow that takes place before this 
condition of equilibrium is reached is a measure of the constant-volume 
heat of reaction. A corresponding constant-pressure heat of reaction, based 
on the attainment of pressure as well as temperature equilibrium with the 
environment, is also of importance, but its discussion will be left for later 
coverage. It will be noted that, in order to bring the products back to 
the same temperature as the reactants, heat must leave the system— 
negative heat when the conventions of engineering thermodynamics are 
applied. In the equation as the chemist writes it, this heat is given a 
positive sign; the convention in regard to the sign of heat flow is thus, in 
the field of chemical thermodynamics, opposite to that which applies in 
engineering thermodynamics. 

Once more for ready reference, when we wish to call attention to the 
possibility of the components of a system releasing energy through a 
chemical reaction such as that described above, we shall speak of the 
energy so released as chemical energy. Chemical energy is, of course, to be 
included in an inventory of stored energy, but as long as there is no change 
in the chemical form of the system, it can be disregarded in calculating 
the total change of stored energy of the system. 

Energy may be stored in forms other than those discussed above. 
These would include magnetism, electricity, and capillarity. These 
forms of stored energy will not be discussed in detail at this time since 
they are of minor importance in engineering thermodynamics. Mag¬ 
netism and electricity are both basically ascribed to the alignment of the 
positive and negative charges characteristic of the structure of matter. 
Capillarity is the ability of a liquid in a small tube to maintain a level 
above or below that of the liquid which surrounds the tube; it is due to 
that property of liquids which is called surface tension. It is because of 
surface tension that liquids can maintain a free surface. 

1:4. Properties of the System. A state of a given system is its condi¬ 
tion or its position with respect to other systems. A state is identified in 
terms of the properties of the system. Typical properties are the pres¬ 
sure, the volume, the temperature, the velocity, and the elevation of the 
system. All of these are, without doubt, observable characteristics which 
can be used to describe the state of the system at the instant of observa¬ 
tion. The various forms of stored energy that have been discussed in the 
preceding article are, it would seem, also properties of the system since 
their magnitudes depend on characteristics of the system, like tempera¬ 
ture, which have fixed values depending on the state. On the other hand, 
heat and work could not be classified as properties since they are forms in 


8 


BASIC ENGINEERING THERMODYNAMICS 


which energy crosses the boundaries of the system; as a result of their 
passage a change of state could be expected. A change of state may take 
place, however, even if no work or heat flow are involved. 

Pressure is the force exerted by a system on a unit area of its boundary. 
On the basis of the kinetic theory of matter it may be interpreted as the 
momentum of the molecules which strike that portion of the boundary in 
unit time. It is proportional to their mass and their velocity. Because 
of the very large number of these molecules it would seem that, for a 
finite system of uniform composition and uniform rate of molecular move¬ 
ment (temperature), the law of averages would assure that the pressure 
would be uniform over all segments of the boundary. This would be true 
except that, under the influence of gravity, the molecules above press 
down on those in the lower part of the system, compacting them so that a 
greater number strike a unit boundary area in unit time. The difference 
in pressure so created is quite small in the low-density systems of limited 
height that are of principal concern to the engineer, and it is the usual 
practice to neglect this effect of gravity in engineering thermodynamics. 

Pressure is measured in units which are consistent with its definition as 
force per unit area, as pounds per square inch or pounds per square foot, 
for example. In this text these will be the units ordinarily employed, the 
symbol P being used if measurement in terms of pounds per square foot 
is intended, p for pounds per square inch. Pressure is sometimes con¬ 
veniently measured in terms of the height of some fluid such as mercury; 
the pressure indicated in this case is that necessary to support a column 
of the fluid of the given height under the influence of gravity. The 
equivalent pressure is the product of the density and the height, or 
P = wz. The pressure equivalent to a column of mercury one inch in 
height is 0.491 psi. 

The nomenclature of pressure includes the terms absolute zero of pres¬ 
sure, atmospheric pressure, gage pressure, absolute pressure, and vacuum. 
Absolute zero of pressure is found in a space in which there is no molecular 
momentum, owing either to the absence of molecules or to the absence of 
molecular motion; it is therefore characteristic of a void and may be 
thought of as the pressure at the top of the earth’s atmosphere. Atmos¬ 
pheric pressure is due to the weight of the earth’s atmosphere and varies 
with altitude and with weather conditions. Standard atmospheric pres¬ 
sure is equivalent to a barometric height of 29.92 in. Hg (or 14.7 psi) and 
represents the average atmospheric pressure at sea level. 

Pressure-measuring devices measure differences in pressure and, in 
engineering practice, frequently the difference between the pressure of the 
system and that of the atmosphere. A pressure gage on a boiler, for 
example, gives the difference between the pressure of the steam on the 
inside and the atmosphere on the outside of the boiler, and this difference 


FUNDAMENTAL DEFINITIONS AND CONCEPTS 


9 


is called the gage pressure. Gage pressure may be defined as the pressure 
in excess of that of the atmosphere, and the corresponding absolute pres¬ 
sure may be obtained by adding the atmospheric pressure to the gage 
pressure. A pressure less than atmospheric pressure is called a vacuum 
and may be recorded on a gage or in terms of the height of a fluid column. 
Vacuum is the amount by which the pressure is less than the pressure of 
the surrounding atmosphere; the corresponding absolute pressure is 
lound by subtracting the vacuum from the atmospheric pressure. 

The temperature of a system measures its capacity for transferring heat 
to other systems. If the temperature of those external systems becomes 
equal to that of the system, the transfer of heat ceases; this does not, 
however, preclude the transfer of energy in the form of work. Based 
upon the kinetic theory of matter and the method by which energy is 
stored in the molecule (see Art. 1:3), temperature may be defined as a 
measure of the kinetic energy of the molecule. In the case of gases at 
low pressure, for which changes in potential molecular energy are negligi¬ 
ble, the change of temperature may account for the total change in the 
stored energy of the molecules of which the system is composed. It 
would not account, however, for changes in other forms of stored energy 
such as stored mechanical energy. Nor would a unit change in tempera¬ 
ture necessarily represent the same change in stored kinetic molecular 
energy at all temperature levels. 

A rise in the temperature of a system is an unfailing indication that its 
molecules have increased velocity, but equality of temperature between 
two bodies, as evidenced by the fact that there is no change in any 
observable characteristic when they are placed in communication such as 
would result from a flow of heat, does not necessarily mean that the 
velocity of their molecules is the same unless the molecules of the two 
systems have equal mass. 

Temperatures are often measured by bringing a carefully calibrated 
external system into communication with the system whose temperature 
is to be ascertained and allowing equilibrium to be established between 
the two. The effect on the properties of the calibrated system is then 
observed and interpreted in terms of the common temperature of the two 
systems. The effects that are commonly used for the measurement of 
temperature in these calibrated systems, which have the general name of 
thermometers , include the change in volume of a solid, a liquid, or a gas 
while the pressure is constant, the change in pressure of a gas at constant 
volume, the change in vapor pressure of liquids, and the change in elec¬ 
trical resistance. Other methods are based on the thermoelectric proper¬ 
ties (the principle of the thermal couple) or on the amount of radiant 
energy emitted (the optical pyrometer). No one method of temperature 
measurement is practical over the complete range of temperatures. 


10 


BASIC ENGINEERING THERMODYNAMICS 


Scales of temperature may be entirely arbitrary. Two scales in com¬ 
mon use are the centigrade and the Fahrenheit. On any scale of tempera¬ 
ture there are certain “anchor points” which are used to orient a reading 
on that scale and make possible its comparison with standard levels of 
molecular activity and with temperatures on other scales. On the centi- 
grade scale the zero point corresponds to the temperature of equilibrium 
between ice and water under a pressure of 1 standard atmosphere (the 
melting temperature of ice at that pressure), and a temperature of 100 
deg on the centigrade scale denotes a temperature level at which water 
and steam are in equilibrium under the same pressure of 1 standard 
atmosphere (the boiling temperature of water at that pressure). There 
is thus a spread of 100 deg of centigrade temperature between these two 
points. On the Fahrenheit scale the first of these points is assigned a 
temperature of 32 deg, the second falls at 212 deg, a temperature differ¬ 
ence of 180 Fahrenheit degrees. Thus the change of temperature repre¬ 
sented by 1 deg on the centigrade scale is equivalent to that denoted by 
f°F. For conversion of temperatures on either of these two scales to 
equivalent temperatures on the other, the following formulas will be 
found convenient : 

Degrees centigrade = f(degrees Fahrenheit — 32) (1:1) 

Degrees Fahrenheit = f(degrees centigrade) + 32 (1:2) 

For the purpose of measuring temperatures which are well above or 
below the range of ordinary atmospheric temperatures, other anchor 
points are desirable. These were selected in 1927 as the result of inter¬ 
national agreement and form the basis of the international scale of tem¬ 
perature. In addition to the two that were cited above, international 
recognition was given to the boiling point of oxygen ( — 182.97°C), the 
boiling point of liquid sulfur (444.60°C), the melting point of solid silver 
(960.5°C), and the melting point of solid gold (1063°C); all of these points 
correspond to a pressure of 1 standard atmosphere. 

Standard methods of evaluating intermediate temperatures were also 
adopted by action taken at this international conference. These methods 
are based on the use of electrical-resistance devices up to a temperature of 
660°C, the use of a thermal couple between that temperature and the gold 
point (1063°C), and, above the gold point, by means of an optical 
pyrometer. 

The zero points on the centigrade and Fahrenheit scales of temperature 
have been selected arbitrarily; temperatures may be demonstrated that 
are represented as negative temperatures on both scales. As will be 
shown later, there is a level of temperature below which it is impossible to 
go; this temperature would correspond, according to the molecular theory, 
to an absence of molecular movement. This level is called absolute zero 


FUNDAMENTAL DEFINITIONS AND CONCEPTS 


11 


of temperature. On the centigrade scale it is located at —273.16 deg, on 
the Fahrenheit at —459.69 deg. These values are approximate and must 
be deduced since, again as will be shown later, absolute zero of tempera¬ 
ture cannot be attained. The boiling point of water at standard atmos¬ 
pheric pressure would thus be located at 373.16 deg on a scale of tempera¬ 
ture that had its zero point at absolute zero and was laid off in centigrade 
degrees. The same point would be represented by a temperature of 
671.69 deg on a similar scale of temperature which was subdivided into 
Fahrenheit degrees. Both of these are called absolute scales of tempera¬ 
ture, and temperatures as measured on them are called absolute tempera¬ 
tures. It is customary to refer to absolute temperatures measured in 
centigrade degrees as degrees Kelvin (°K) and, as measured in Fahrenheit 
degrees, as degrees Rankine (°R). Scalar temperatures on either the 
centigrade or the Fahrenheit scale will be denoted by the symbol t, abso¬ 
lute temperatures by T. Thus the relation between degrees Rankine and 
degrees Fahrenheit is given by 

T = t + 459.69 

In slide-rule calculations it is usual to substitute the value 460 for 459.69 
in this equation. 

The total volume ( V ) of a system is the space that it occupies; the 
specific volume (v) is the space occupied by a unit mass. When the 
specific volume varies throughout a system, as is the case under the influ¬ 
ence of gravity or when the system is a mixture of phases, its value must 
be defined, with respect to a given point in the system, as the space 
occupied by an infinitesimal part of the system divided by the mass of 
that infinitesimal part. Density is the reciprocal of specific volume. 

In the course of the development of thermodynamic theory it will be 
necessary and desirable to introduce properties other than those which 
have been discussed in the preceding paragraphs. Some others which 
will occur to the reader are elevation, velocity, electrical resistance, and 
viscosity. The pressure, volume, and temperature are the properties 
that are most commonly used in describing the state of the system and 
might be called the primary properties. Equations of state for the various 
materials of which systems are composed ordinarily show the relationship 
between these primary properties. 

Properties may be classified as extensive and intensive. The value of an 
extensive property which applies to a given system depends on the mass 
of that system; the total volume and the total stored energy of the system 
are examples of extensive properties. Pressure, temperature, elevation, 
velocity, electrical resistance, and viscosity are intensive properties; their 
values for the system as a whole are not the sums of the corresponding 
values for the various parts of which the system is composed. 


12 


BASIC ENGINEERING THERMODYNAMICS 


The identification of a quantity which is associated with a system as a 
property of that system is often important. For this purpose it will be 
remembered that, whenever a system returns to a given state, all of its 
properties must have their original values. Furthermore, when a system 
changes its state, the change of any property will depend only on the end 
states and not at all upon the means by which the change of state was 
accomplished. 

1:5. The Pure Substance and the Simple System. The pure substance 
is homogeneous in composition and in chemical form and retains its 
chemical identity. A mixture of liquid water and steam would satisfy 
these requirements and would be classed as a pure substance even though 
the density is not uniform throughout the mixture. Dry air would 
ordinarily be a pure substance although it is a mixture of gases. How¬ 
ever, at temperatures low enough to bring about a partial condensation of 
its constituents, it would be found that the percentage of the more volatile 
components would be lower in the liquid than in the gases of the resulting 
two-phase mixture; as a mixture of two phases, air is not a pure substance. 
Atmospheric air is a pure substance only as long as there is no condensa¬ 
tion of its water vapor content. A mixture of hydrogen and oxygen in 
the gaseous phase is classified as a pure substance; when the two gases 
unite chemically to form water vapor, a different pure substance results. 

The importance of the thermodynamic concept of a pure substance lies 
in the fact that, in the absence of motion, gravity, capillarity, magnetism, 
and electricity, all properties of a system composed of the substance are 
established if only two, which are independent of each other, are known. 
If these two properties are used as the coordinates of a graph and if the 
known values of the two properties that correspond to a given state of the 
pure substance are used to locate a point on the graph, the position of that 
point also fixes the values of all other properties associated with that 
state. Thus it is possible to write an equation of state for the pure sub¬ 
stance which will express any property in terms of not more than two 
other (and independent) properties. A partial list of the properties for 
which such equations can be written includes pressure, temperature, 
specific volume, stored molecular energy, density, viscosity, and electrical 
resistivity. Many others will later be added to this list. 

It is noted that the two properties which are to be the coordinates of 
the graph must be independent of each other. The pressure and the 
volume, for instance, are often used and have a special usefulness to be 
developed later. Mutually dependent properties, such as specific volume 
and density, could not be used for obvious reasons. For mixtures of 
phases, the pressure and temperature of the pure substance are interde¬ 
pendent and could not be used although, for any single phase, they would 
serve. 


FUNDAMENTAL DEFINITIONS AND CONCEPTS 


13 


The only form of stored energy of which the relative value could be 
expressed in the form of an equation of state for a given pure substance is 
stored molecular energy, to which we shall hereafter apply the name 
internal energy and which we shall designate by the symbol U for total, u 
for specific internal energy. The attention of the reader is called to the 
fact that all forms of stored energy, including the kinetic and potential 
energy of the system as a whole (stored mechanical energy), are, in the 
broad sense, internal energy and that that term (though not the symbol 
U) as used in many texts includes them. In this book, for purposes of 
clarity, the sum of all forms of stored energy will be called stored energy 


i 


2 


/ 


2 

/ 


V \/ 

(a) (b) 

Fig. 1:1. The constant-volume path. 


and will be represented by the symbol E for total, e for specific stored 
energy. These symbols are those conventionally employed to designate 
stored energy in general. 

The simplest type of system with which it is possible to deal consists of 
a pure substance, the effects of motion, gravity, capillarity, magnetism, 
and electricity being either absent or negligible in amount. Since it is 
changes in the amount of stored energy that are of principal concern and 
since, by definition of the pure substance, the stored energy due to chemi¬ 
cal form does not change, Ae = Au. Indeed, the forms of stored energy 
that are either present in negligible amount or do not change in amount 
as the state of the system changes are usually ignored, and the internal 
energy represents the total stored energy, or e = u. 

The system is only slightly less simple when the effects of motion and 
gravity are included, as is often the case for open-system analysis. In 
this case, e may be obtained by adding to u terms which represent the 
amount of additional energy storage due to these two effects. 

1:6. The state path is the succession of states through which the system 
passes as it changes its state. Consider a simple gas system confined in a 
closed rigid shell. The condition of this system originally is represented 
by the position of point 1 on Fig. 1:1a; here the coordinates are tempera- 








14 


BASIC ENGINEERING THERMODYNAMICS 


ture and volume. If heat is supplied to the system and passes inward 
across its boundaries, the temperature will rise and the path 1-2 will be 
traced. Any point on this path will represent an intermediate state of 
the system at which not only the temperature and volume but all proper¬ 
ties of the system will have fixed values. Thus the same state path may 
be represented, as in Fig. 1:16, on coordinates of pressure and volume; in 
fact, any pair of independent properties of the system could be utilized 
for this purpose. 

The path followed in this example is described as a constant-volume 
path. Other special kinds of state paths which will suggest themselves 
to the reader on the basis of properties already defined are those of con- 




Fig. 1:2. Graphical description of a Fig. 1:3. The paddle-wheel process, 
state path. 

stant pressure, constant temperature, and constant internal energy. 
Whenever any property is constant during a change in state and the end 
states are known, the state path for a given simple system is completely 
defined for any pair of coordinates that represent independent properties 
of the system. The same is true, of course, when no property is constant 
but the path is described in terms of the change in properties. The 
description in this case may be graphical as indicated in Fig. 1:2. 

1:7. The Process. A description of the thermodynamic process 
involves not only a description of the path but also the method by which 
the path was followed. For example, the path illustrated in Fig. 1:1 was 
ascribed as due to the passage of heat across the system boundaries. The 
same path could have been followed if the energy that crossed the bounda¬ 
ries of the system had been in the form of work. This possibility is 
illustrated in Fig. 1:3, in which a paddle wheel, driven by an external 
agency, stirs the confined system and thereby increases its temperature; 
this is a variation of a fundamental experiment in elementary physics. 



























FUNDAMENTAL DEFINITIONS AND CONCEPTS 


15 


Thus the same path may be followed during either a no-work or a no-heat 
process. An infinite variety of heat-work combinations are possible for 
every state path. 

Processes which take place without heat flow are called adiabatics. 
The description of a process as an adiabatic does not define the path, even 
if the end states are known, unless some hint is given as to the rate at 
which work crosses the boundaries of the system during the process. 

1:8. The Cycle. Any path or process whose end states are identical is 
called a cycle. However, the description of a cycle is not thermodynami¬ 
cally complete unless it is described in 
terms of the process or processes trav¬ 
ersed during the cycle. In Fig. 1:1 a 
cycle is formed if the state of the system 
is returned to that represented by point 
1. This may be done by inducing heat 
to flow out of the system; the path 
followed during the original process is 
exactly retraced, but if the increase of 
system temperature during the process 
1-2 was due to work supplied the sys¬ 
tem rather than heat flow, the same can¬ 
not be said of the process. Note also 



V 


Fig. 1:4. The cycle. 


that whereas the path 1-2 can be followed without any heat flow what¬ 
ever the return passage, 2-1, must involve the flow of heat; the paddle- 
wheel process cannot be reversed to cause work alone to flow out of the 
system. 

If the path 1-2 in Fig. 1:1 represents a process involving no work, the 
heat flow outward during the return to state 1 is equal to the heat 
received as the temperature of the system increased; the net heat flow 
for the cycle is thus zero or f dQ = 0. On the other hand, if heat flow 
is involved only in the return process, f dQ <0. Similarly as regards 
the net work of the cycle, in the first instance f dW = 0 and, in the 
second, f dW < 0. 

It is not intended to imply that the cycle always consists of a direct 
reversal of a state path; the more usual (and more useful) form of cycle 
has a state path which encloses an area as shown in Fig. 1:4. 

1:9. Identification of a Property. It is sometimes necessary to estab¬ 
lish whether a quantity associated with a given system is a property of 
that system. If it is a property, it may be used as a coordinate in plotting 
a state or a path of the system and its change may be computed for any 
process as based on the end states alone without regard for the manner in 
which the change of state was effected. It is this characteristic of a 
property which makes possible its identification. 




16 


BASIC ENGINEERING THERMODYNAMICS 


In traversing a cycle a return is always made to the starting point. 
This is an identical state with identical properties, and therefore, if it can 
be shown that the total change of the quantity around any and all cycles 
which the system can be caused to traverse is zero, that quantity is 
established as a property of the system. Conversely, if any cycle can be 
found for which the line integral of the quantity is not equal to zero, it 
cannot be classified as a property. A brief review of Art. 1:8 will indicate 
that heat and work, at least as they are considered individually, are not 
properties. 

If a quantity is expressed entirely in terms of the whole value (rather 
than the changes) of known properties of the system, it is identifiable as a 
property of that system; such properties are called composite properties. 

It will be remembered that a simple system has only two independent 
properties and that any third property may be expressed in terms of these 
two. The equation that connects the three represents a three-dimensional 
surface for which the coordinates are, respectively, the three properties. 
Thus, if x and y are the independent properties and P is the dependent 
property, 

P = f(x,y) (1:3) 


dP = ^ dx + dy = M dx + N dy 
dx dy 


(1:4) 


If P is a function of x and y, as it must be if it is a property, its second 
derivative with respect to x and y is 


d 2 P 




dx dy dy \dx 



or 


dM dN 


dy 


dx 


(1:5) 


That is, the order of differentiation is unimportant. If this requirement 
is satisfied, M dx + N dy is called an exact differential. If, on the other 
hand, P is not a property (though its amount may be expressed in the 


form M dx + N dy) is not equal to ^ 


This furnishes another 


dy 1 dx 

means by which a property may be identified. 

1:10. The Equation of State. The substance of which the simple 
system is composed retains its chemical identity, and, as has been previ¬ 
ously stated, it has, in the absence of motion, gravity, capillarity, mag¬ 
netism, and electricity, only two independent properties. For ease of 
reference, it will be called a pure substance though the use of that term 
should not be understood to imply that the pure substance may not con¬ 
sist of a mixture of component substances. However, if a mixture, the 
proportions of its components must be uniform throughout the pure sub¬ 
stance. Any property of the pure substance may be expressed in terms of 
not more than two other properties, and the equation which results will 







FUNDAMENTAL DEFINITIONS AND CONCEPTS 


17 


represent a surface. This equation, it will be remembered, has been 
called an equation of state. One of the simplest examples of an equation 
of state is that which applies to an ideal gas and connects the pressure, 
volume, and temperature; it will be developed in a later chapter but is 
introduced at this point to aid our discussion. 

pv = CT (1:6) 

in which C is a constant that varies with the gas and p is the absolute pres¬ 
sure. That C is not a dimensionless constant will be evident if the dimen¬ 
sions of the product pv are compared with those of T; its value will change 
if, for instance, the pressure unit is changed from pounds per square inch 
absolute to pounds per square foot absolute. Equations of state are 
ordinarily much more complicated in their form than Eq. (1:6), and it is 
seldom possible to write the same equation to fit both of two phases 
accurately. That need not concern us for the present; more general 
equations of state will be discussed in a later chapter. 

Equation (1:6) is only one of a number of equations of state which 
apply to the ideal gas; it connects the primary properties. It will be 
shown later that another relationship, based directly upon and following 
from Eq. (1:6), is 


u = A Bpv (1:7) 

in which A and B are constants though, since they are not dimensionless, 
they will change in value for a given pure substance as u, p, and v are 
expressed in varying units. Again it should be emphasized that this is 
not a general form for the calculation of the internal energy of all pure 
substances. 


Example 1:10. For a certain pure substance, Eqs. (1:6) and (1:7) are adequate as 
the corresponding equations of state if A. =16 , B = 0.46, and C = 0.37 when p is 
measured in pounds per square inch absolute, v in cubic feet per pound, T in degrees 
Rankine, and u in Btu per pound. (The Btu, or British thermal unit, is a unit of 
energy that will later be defined. It approximates in magnitude 778 ft-lb.) A system 
consisting of 3 lb of this substance is confined in a rigid shell at a pressure of 50 psig 
and a temperature of 60°F. Heat is supplied the system, and, as a result, its tem¬ 
perature increases to 120°F. Calculate (a) the internal volume of the container, 
(b) the final pressure of the system, and (c) the change of internal energy which 
resulted from the process. 

Solution: 


(a) v i 
Vi 

(b ) Vi 

V 2 


CTi 0.37(460 + 60) 


= 2.97 ft 3 /lb 


P x 50 + 14.7 

Mv ! = (3) (2.97) = 8.91 ft 3 

Vx = 2.97 ft 3 /lb; T 2 = 120 + 460 = 580°R 


CT 2 (0.37) (580) 


v 2 


= 72.2 psia or 57.5 psig 


2.97 






18 


BASIC ENGINEERING THERMODYNAMICS 


(c) u x = A + BpxVx = 16 + (0.46) (64.7) (2.97) = 104.5 Btu/lb 

u 2 = A + Bp 2 v 2 = 16 + (0.46) (72.2) (2.97) = 114.7 Btu/lb 

u 2 - ux = 114.7 - 104.5 = 10.2 Btu/lb 
U t ~ Ux = M(ui - ux) = (3) (10.2) = 30.6 Btu 

Problems 

1. Classify any of the following which are properties as intensive or extensive; 

work, kinetic mechanical energy, modulus of elasticity, mass, moment of inertia, 

coefficient of linear expansion, color, stress (in a solid system), surface tension (in a 
liquid system). 

2. A ton of gold is shipped from Alaska, where g = 32.22, to San Francisco, where 
g = 32.15. What is its weight in San Francisco? What is the mass of the gold in 
Alaska? In San Francisco? 

3. Convert the following pressures to pounds per square inch absolute. Barometer 
= 29 in. Hg (a) 3000 psfa; (6) 25 psig; (c) 27 in. Hg vacuum; (d) 2000 psfg. 

4. Convert the following pressures to pounds per square foot gage. Barometer 
= 30 in. Hg. (a) 1400 psig; (6) 1000 psfa; (c) 20 in. Hg vacuum; (d) 150 psia. 

5. Change the following temperatures to degrees Rankine: (a) 400°F; ( b ) — 5°C; 
(c) 220°K. 

6. Change the following temperatures to degrees Fahrenheit: (a) 440°R; ( b ) 15°C; 
(c) 260°K. 

7. Devise a means of measuring temperature (a thermometer) which employs 
principles not ordinarily made use of in thermometers. Explain how you would 
calibrate your device. 

8. Five pounds of air occupies a total volume of 60 cu ft. What is the weight 
density of the air? Its mass density in slugs per cubic foot? 

9. In the description of each of the following processses, the system is defined by 
the italicized words. In each case (1) state whether the system is an open or a closed 
system and (2) whether it is a simple system. (3) Determine whether the heat flow 
that accompanies the process is positive, negative, or zero. (4) Do the same for the 
work of the process. 

(a) A container with rigid nonconducting walls holds air and an electric heating ele¬ 
ment, which is connected to a source of power external to the container. The tem¬ 
perature and pressure of the air increase. 

( b ) A container with rigid nonconducting walls holds air and an electric heating 
element which is connected to a source of power external to the container. The tem¬ 
perature and pressure of the air increase. 

(c) A container with rigid nonconducting walls holds air and an electric heating 
element with connected storage battery. The temperature and pressure of the air 
increase. 

(d) A teakettle with its contents of water, water vapor , and air is placed on a stove. 
As the result, steam at atmospheric pressure issues from the spout. 

(e) A vessel with rigid nonconducting walls is divided into two compartments by a 
rigid nonconducting partition. One of the compartments is filled with air; the other 
is a void, containing no matter. The partition is removed, and the air expands to fill 
the larger space. 

(/) Air is confined in a cylinder with rigid nonconducting walls and head but fitted 
with a gastight piston, also nonconducting. The pressure of the air inside the cylinder 
is less than that on the outside face of the piston, and the piston moves, compressing 
the charge of air. 

(g) Steam flows through a turbine with nonconducting walls, turning the turbine 
shaft against the resistance offered by an electric generator. 


FUNDAMENTAL DEFINITIONS AND CONCEPTS 


19 


C h) Water enters a boiler, and steam leaves it, both at the same pressure; the specific 
volume of the steam is greater than that of the water. 

(i) A mixture of hydrogen and oxygen at atmospheric pressure and temperature in 
the proper proportions for complete combustion is confined within a rigid nonconduct¬ 
ing pressure vessel. A spark, which may be considered negligible, ignites the mixture. 

(j) Same as (i) except that the walls of the vessel, though rigid, are conducting. The 
process ends when temperature equilibrium with the atmosphere is again established. 

( k ) Same as ( i ) except that the walls of the container are both flexible and conduct¬ 
ing; they do not rupture. The process ends when temperature and pressure equilib¬ 
rium with the atmosphere is again restored. 

10. A closed system traces a cycle of operations. Which of the following are 
necessarily true statements? (a) jfdQ = 0; (6) fdQ >0; (c) jfdQ < 0; (d) j>dP = 0; 
(e) fdP >0; (/) fdP <0; (g) fdW = 0; (h) fdW > 0. 

11. In the following expressions, P, v, s, and T are properties, and a, b, n, R, J , and 
C are constants. In which cases is A a property of the system? When X is found to 
be a property, write an expression for it in terms of other properties and the appropriate 
constants. 


(a) dX i = T ds 
0 b ) dX 2 
(c) dX, 


Pdv 
J 

Pv n ~ l dv + v n dP 
C dT CT 


(d) dX 4 

(e) dX 5 


v 

R dT 
v — b 
R dT 


dv 

RT 


+ 


(v 

RT 


by 


dv 



At the end of the earlier chapters of this text, a number of problems will be stated the 
solutions of which will be based on the use of the equation of state of an ideal gas, following 
the form of Eqs. (1:6) and (1:7). Four gases will be used, and these, for convenience, will 
be designated as gas W, gas X, gas Y, and gas Z. In the table belo w the values of the con¬ 
stants A, B, ana C in Eqs. (1:6) and (1:7) are tabulated for each. The stated values of the 
constants as given in the table are based on p in pounds per square inch absolute, v in 
cubic feet per pound, T in degrees Rankine, and u in Btu per pound. It will be observed 
that gas Y is the gas used in Example 1:10. 


Const 

Gas W 

Gas X 

Gas Y 

Gas Z 

A 

14.7 

854 

16.0 

71.0 

B 

0.639 

0.570 

0.460 

0.282 

C 

0.244 

0.596 

0.370 

2.68 


12 . At standard atmospheric pressure and a temperature of 32°F, what is the specific 
volume of (a) gas W; (b) gas X; (c ) gas Y; (d ) gas Z? 

13. The density of a gas is 0.1 lb/ft 3 at a temperature of 50°F. Under what pres¬ 
sure is it confined if it is (a) gas W ; (6) gas X; (c) gas Y ; (d) gas Z? 

14. A gas is confined under a pressure of 5 atm abs, and its specific volume is 5 ft 3 /lb. 
What is its temperature in degrees Fahrenheit if it is (a) gas W ; (6) gas X; (c) gas Y; 
(d) gas Z? 





















20 


BASIC ENGINEERING THERMODYNAMICS 


15. What is the internal energy of a 5-lb gas system at atmospheric pressure and a 
temperature of 100°F if it is composed of (a) gas W; (b ) gas X; (c) gas Y ; ( d ) gas Z? 

16. What is the rate of change of specific internal energy with respect to temperature 
for (a) gas W; (b ) gas X; (c) gas Y; (d ) gas Z? 

17. Show that the coefficient of cubical expansion (the rate of increase of volume per 
unit increase in temperature) at constant pressure is the same for all four gases at the 
same temperature. What is its value at 100°F? 

18. Assume that the state path shown in Fig. 1:2 represents an isothermal (constant- 
temperature) expansion and that p i = 120 psia, t\ = 300°F, and V\ = 1 ft 3 . What 
weight of gas takes part in the expansion if it is (a) gas W; ( b) gas X; (c) gas F; ( d ) 
gas Z? Write the equation of the state path, and show that it will be the same for all 
of the gases. 

19. One pound of a gas begins an isothermal expansion at pi = 120 psia, t x = 300°F. 
Write the equation of the state path if the gas is (a) gas W, ( b ) gas X; (c) gas Y; ( d ) 
gas Z. Show the relative position of these state paths on the same pv diagram. 


A, B, C 
e 
E 

9 

m 

M 

V 
P 

Q 

t 

T 

u 

U 

v 

V 
w 
W 

x, y 

z 


Symbols 

constants 

stored energy of a system of unit mass in general 

stored energy of a system in general 

acceleration due to gravity 

mass, slugs 

mass, lb; also, weight 

pressure, psi 

pressure, psf; also, any dependent property 
heat flow 

scalar temperature 
absolute temperature 

energy stored as thermal energy in a system of unit mass 

stored thermal energy of a system 

volume of a system of unit mass, specific volume 

volume of a system 

density 

work 

any two independent properties 
height, elevation above a reference level 


CHAPTER 2 


THE FIRST LAW AND THE CLOSED SYSTEM 

2:1. The First Law of thermodynamics is a statement of the implica¬ 
tions of the law of conservation of energy in a form that is specifically 
designed for direct application to the development of thermodynamic 
theory. It may be stated, in one of its many variations, as follows: 

When any system is carried through a process or series of processes that 
eventually return it to its original state , the net amount of heat flow into the 
system is equal to the net amount of work delivered to external systems during 
the cycle. 

The only proof of this law is the failure of all attempts to prove it false. 
Because of those failures we have confidence in its validity and are content 
to use it as a part of the foundation on which the structure of thermo¬ 
dynamic theory will be built. 

The First Law makes some progress in that direction immediately 
possible since, taken in conjunction with its parent law, the law of con¬ 
servation of energy, it confirms the concept of stored energy as a property 
of the system. For when the system has returned to an identical state at 
which it must have identical properties, as was the result after its traversal 
of a cycle, the First Law states that in passage around that cycle no net 
energy could have been transferred to or from the system. Since energy 
can neither be created nor destroyed, the amount of energy stored in the 
system must be the same at the end of the cycle as at the beginning. 
This is an application of one of the methods suggested for the identifica¬ 
tion of a property in Art. 1:9. 

The various forms of energy are ordinarily measured in different units. 
For instance, we usually contemplate the measurement of heat in British 
thermal units, of work in foot-pounds; in tables of properties, internal 
energy is stated conventionally in Btu. Interpreting the meaning of the 
First Law as it applies to these units of energy, it is seen that a fixed 
relationship must exist between them. Stated in somewhat different 
fashion, it is entirely practicable to express heat flow quantitatively in 
terms of foot-pounds and work in Btu. 

The British thermal unit (Btu) was originally defined as xio- of the 
amount of heat necessary to raise the temperature of one pound of water 
from the freezing point (32°F) to the boiling point (212°F) under a con¬ 
stant pressure of one atmosphere. The same international conference 

21 


22 


BASIC ENGINEERING THERMODYNAMICS 


that established the international scale of temperature to which reference 
was made in the preceding chapter, recognizing that the measurement of 
heat was inherently less precise than the measurement of work, redefined 
the Btu as the equivalent of 778.2 ft-lb of energy. This ratio corresponds, 
according to recent experimental data, to a unit of energy very slightly 
smaller than was described in the original definition of the Btu. 

It will be our policy to measure heat flow and internal energy in Btu 
and to measure work in foot-pounds. The relation between these units 
is 1 Btu = 778.2 ft-lb and will be symbolized by a proportionality factor 
J( — 778, approximately). Thus the First Law, as stated above, may be 
represented by the mathematical equation : 

jfdQ = fdW (2:1) 

2:2. The Closed-system Process. Let a closed system be caused to 
follow any state path during a change from state 1 to state 2. The change 
in stored energy, now established as a property of the system, will be 
dependent only on the end states and may be designated as E 2 — E h or 
as A E. But the change in E must have resulted from a net flow of 
energy across the system boundaries, and, for a closed system, this could 
have been only in the form of heat or work. The net amount of heat flow 
which accompanied the process is dependent not only upon the path but 
upon the way in which that path was followed. To represent it we shall 
use the notation iQ 2 , indicating that this heat flow took place as the 
system changed its state, depends on the manner in which the change of 
state was effected, and cannot be calculated on the basis of the end states 
alone. Similar reasoning may be applied to the net work that accom¬ 
panied the process, and it will be designated as AV 2 . Then 

Energy flow in—energy flow out = change in stored energy 

or, taking into account the conventions that have been adopted as to the 
signs of heat and work, 

J 1 Q 2 — ill 2 = J{E 2 — Ei) = J A E (2:2) 

This is the general statement of the First Law as it applies to a closed- 
system process. In the absence of motion, gravity, capillarity, magne¬ 
tism, and electricity, the foregoing equation becomes 

J 1 Q 2 — iW 2 — J(JJ 2 — U 1 ) = J A l (2:3) 

Occasionally the process that connects states 1 and 2 is so complicated 
that it is impossible to fit a single description to its entire length. In that 
case it may be divided into segments and described piece by piece. The 
change of stored energy over an infinitesimal segment may be expressed 
as dE (or dU in the case of the simple system) and the total of all these 
infinitesimal changes summed to measure the total change of stored 


THE FIRST LAW AND THE CLOSED SYSTEM 


23 


energy of the system during the entire process, or f* dE = E 2 — E\. 

The same general procedure can be followed in measuring iQ 2 and \W 2 
except that the infinitesimal amount of heat and work, not being proper¬ 
ties and therefore not dependent solely on the end states of the infini¬ 
tesimal segments into which the process has been divided, cannot properly 
be expressed as exact derivatives. To indicate this difference and to 
keep ourselves reminded of it, we shall employ the symbols dQ and 

dTT to represent their infinitesimal magnitudes. Then 1 Q 2 = dQ, and 
iW 2 = J 1 dW. For an infinitesimal process, Eqs. (2:2) and (2:3) become 

J dQ — dW = J dE ■ (general, closed system) (2:4) 

J dQ — dW = J dlJ (simple closed system) (2:5) 

Examination of Eqs. (2:2) to (2:5) indicates that, while heat and work 
considered individually are not properties of the closed system, their 
difference during any process through which the system is carried is a 
property of the system since it will depend only on the end states. Thus 
if by some means (as yet unknown to us) the maximum amount of work 
that can accompany the tracing of a given path with fixed end points 
can be established, the maximum amount of heat flow that can be associ¬ 
ated with that path can be calculated when the change of stored energy 

of the system, as based on the end states, may be determined. As the 

work which accompanies the tracing of the given path decreases from this 
maximum, there must be a corresponding and equal reduction in the 
heat flow since the change in stored energy, being dependent only on the 
end states, will remain constant. 

Example 2:2 A. During a certain process a simple system receives 20 Btu of heat 
as its internal energy is increased by 30 Btu. What amount of work is involved in 
the process? 

Solution. Applying Eq. (2:3), 

(778) (+20) - AV 2 = (778) (+30) 

1 W 0 = - 7780 ft-lb 

Justifying the signs used in substituting in the equation, it will be remembered that 
heat flow into the system is positive and that, if the internal energy increases, U 2 > U x . 
Similarly, consistent with the sign convention that applies to work, it is evident that 
work was done on the system during the process. 

Example 2:2 B. The maximum amount of work that can accompany a certain path 
for a simple system is -10,000 ft-lb. The change of internal energy of the system 
between the end states of that path is +20 Btu. (a) What is the maximum heat flow 
that can take place during the traversal of the specified path? ( b ) If, owing to a 
change in the description of the method by which the given path was traversed, the 
amount of work was reduced to —20,000 ft-lb, what heat flow would accompany this 
new process ? 


24 


BASIC ENGINEERING THERMODYNAMICS 


Solution: 

(а) Applying Eq. 2:3, 

778(iQ 2 )max - (-10,000) = (778) (+20) 

( 1 Q 2 )max = +7.15 Btu (into the system) 

(б) Note that the amount of work has decreased in the algebraic sense. Substituting 
in Eq. (2:3), 

778 iQ 2 - (-20,000) = (778)(+20) 

1 Q 2 = —5.7 Btu (out of the system) 

Or, applying the principle that the heat flow must have decreased in exactly the 
amount of decrease of work, 

,Q 2 = 7.15 - 10,000/778 = -5.7 Btu 

2:3. Work and the Closed System. In the absence of motion, gravity, 
capillarity, electricity and magnetism, 1 work may pass across the bounda¬ 
ries of the closed system in only two ways. One of these is the paddle- 
wheel method described in Art. 1:7. This process permits only the 
inward passage of work (negative work), and the process cannot be 
retraced (though the system may be returned to its original state over the 
same path). Paddle-wheel work is possible because of a property of the 
fluid called viscosity and takes place through the agency of friction. 
Intermolecular friction accompanies turbulence such as is caused by the 
paddle wheel. Friction always has the effect of producing a change in the 
state of the system , when work is done on it, which could have been produced 
by a flow of heat. But since work is a much more costly and valuable 
form of energy than heat, frictional processes are wasteful processes 
and to be avoided as far as possible. In general, in developing the theo¬ 
retical principles which limit the performance of engineering apparatus, 
it will be our policy to base our calculations on an assumed absence of 
friction. Thus the performance of the actual machine, in the operation 
of which friction always is a factor, can approach but never reach the 
theoretical standards that apply. 

The only other method by which work may pass across the boundaries 
of the closed system is by a change in the position of those boundaries due 
to an expansion or a contraction of the system. An increase in system 
volume would, in general, indicate positive work, a decrease would be 
accompanied by negative work. Typical of a transfer of energy in the 
form of work in this manner is work done on a piston. 

1 In analyzing thermodynamic processes of the closed system, energy storage due to 
the effects of motion, gravity, capillarity, magnetism, and electricity will hereafter 
be considered negligible unless otherwise specifically stated. Also, a simple system 
will be assumed so that all properties are established if the values of only two which 
are independent of each other are known. 


THE FIRST LAW AND THE CLOSED SYSTEM 


25 


2:4. Maximum Work on a Piston. Suppose the system to be confined 
within a cylinder which is fitted with a movable piston as illustrated in 
Fig. 2:1. The original position of the piston is shown in full lines and the 
corresponding pressure and volume of the system are Pi and V As the 
system expands, overcoming the resistance offered by external systems 
such as the piston, piston rod, fly¬ 
wheel, and any external fluid sys¬ 
tem (as, for instance, the atmos¬ 
phere, which may oppose the 
motion of the piston by exerting 
pressure against its outer surface), 
the variation of system pressure 
and volume is traced, resulting in 
the path 1-2 on the pressure-vol¬ 
ume diagram. In practice, this 
graph may be made by means of an 
instrument called an indicator, 
which is connected, for the meas¬ 
urement of pressure, through one 
of the stationary walls that form a 
part of the system boundary and 
which is indirectly connected to the 
piston to give the lateral movement 
that measures the change in system 
volume. 

As an infinitesimal section of the path is traversed, the average pressure 
is P , and if the area of the face which the piston presents to the system 
(measured in a plane perpendicular to the direction of piston movement) 
is denoted by A, the average force on the piston is PA. During the 
infinitesimal change in volume that has taken place (dV), the piston has 
moved a distance dV/A and the resultant work is (PA){dV/A) = P dV ; 
this is represented by the segmental area shown in Fig. 2:1. For the 
entire process, the work which has left the system is the sum of these areas 
between states 1 and 2, or 

i W t = j*PdV (2:6) 

If an expression for P in terms of V that represents the relation between 
them along the path is available, iW 2 can be found by mathematical 
integration. Or, if more convenient, the area between the path and the 
V axis can be measured graphically or by any other means and iW 2 can be 
thereby evaluated. If i W 2 is to be measured in foot-pounds, that result 
can be obtained by expressing P in pounds per square foot absolute and V 










PQ 




I I 

U 



V, V 2 V 

Fig. 2:1. Maximum work on a piston. 




























26 


BASIC ENGINEERING THERMODYNAMICS 


in cubic feet. Note that as far as the calculation of the work that left the 
system is concerned, it makes no difference whether the resistance which 
was overcome was offered by the piston and other parts of the mechanism 
or by the pressure of the atmosphere operating against the rear face of the 
piston; this explains why P is expressed in term of the absolute pressure. 

In the development of Eq. (2:6) the pressure was explained as that 
operating against any stationary part of the boundary; it is only because 
gravity is neglected that a common pressure can be assigned to parts of 
that boundary at different elevations. During the expansion on which 
our development was based, the piston is moving outward, and the 
velocity with which the molecules of the system strike the inner face of 
the piston must be correspondingly less than their velocity relative to the 
stationary parts of the boundary. Based on the concept of pressure out¬ 
lined in the first chapter, the pressure which the system exerts on the face 
of the piston must be less than that exerted against the stationary bound¬ 
ary and the work accompanying the process less than as calculated by 
Eq. (2:6). Considering the very much greater speed of the molecule as 
compared with piston speeds employed in usual engineering practice, this 
effect is small, but iIT 2 , as calculated by Eq. (2:6), none the less measures 
the maximum work which can accompany the tracing of a given state path 
as the system expands; to develop it in the full amount requires that the 
motion of the piston shall be indefinitely slow. During a compression, 
the piston moves inward, and the pressure that the system applies to the 
inner face of the piston is greater than that against the stationary bounda¬ 
ries. The absolute amount of work involved is, in this case, larger than 
is represented by the area between the path and the V axis. However, 

this is negative work and, in the algebraic sense, smaller than P dV . 

Thus the application of Eq. (2:6) gives the maximum amount of work, 
in the algebraic sense , that can accompany the tracing of the path. This 
is the maximum amount of work to which reference was made in Art. 2:2 

and which, as it now develops, can be measured as f P dV. 

Example 2:4. A closed system changes from an initial pressure of 50 psig and 
volume of 10 ft 3 to a final pressure of 70 psig and a final volume of 8 ft 3 . The path is 
represented by a straight line on a diagram for which the pressure and volume are the 
coordinates, (a) What maximum work could accompany any process during which 
this path was followed? (6) If the original internal energy was 150 Btu and the 
final 170 Btu, what maximum heat flow could have taken place? 

Solution: 

(a) To contrast the two methods by which the work area may be measured, both will 
be applied. Applying Eq. (2:6), 

pi = 64.7 psia; p 2 = 84.7 psia; V\ = 10 ft 3 ; V-i — 8 ft 3 


THE FIRST LAW AND THE CLOSED SYSTEM 


27 


By the methods of analytical geometry, the equation of the straight line passing 
through these points on a pF graph is 

V = 164.7 - 107 or P = 144(164.7 - 107) 

iW, = 144 fC (164.7 - 107 )dV = 144 [ 164.77 - 57*]? 0 = -21,600 ft-lb 

The work area is trapezoidal and is negative in sign since the process proceeds from 

84 7\ 

—-) (-2) = -21,500 ft-lb. 

(6) Maximum heat flow will accompany maximum work, and, applying Eq. (2:3), 

778 i Q 2 - (-21,500) = 778(170 - 150) 

1 Q 2 = —7.6 Btu 

This is the maximum heat flow in the algebraic sense. 

2:5. Useful Work. The reader will have observed that in Fig. 2:1 the 
total work delivered by the system during the expansion process could not 
have been delivered to the more remote parts of the engine mechanism, 
even if all parts had been frictionless, because a portion was dedicated to 
the task of pushing back the atmosphere. This is a characteristic limita¬ 
tion that applies to the process as carried out in practice; the surroundings 
must be taken into consideration. In most cases the surroundings are 
represented by the atmosphere, which is conceived as a medium so vast in 
extent that a change in volume of the finite system produces no significant 
change in the pressure of the medium nor does a flow of heat to or from the 
finite system result in any change in the temperature of the medium. The 
temperature and pressure of the medium are therefore assumed to be con¬ 
stant and are designated as T 0 and P 0 . In the case of the expansion used 
as an example in Art. 2:4 the part of the total work that was absorbed by 
the atmosphere is P 0 (F 2 — VI), and the maximum work which can be 
applied to useful purposes is i!7 2 — Po(V 2 — VI). This is not of immedi¬ 
ate importance to us, but reference will again be made to it at a later stage 
in our discussion. The distinction applies only to the process, for if a 
cycle is eventually traversed, the atmosphere will have contributed as 
much work during the return of the piston to its original position as it 
absorbed on the outward journey. 

2:6. Constant-volume Processes. If a constant-volume path is fol¬ 
lowed, as in Fig. 2:2, the area under the state path, and thus the maximum 
work that can accompany any process which traces that path, is zero. 
As we have seen in Art. 1:7, the path 1-2 could have been traced by sup¬ 
plying work instead of heat to the system (the paddle-wheel process); this 
work would have been negative in sign with respect to the system and 
thus less than zero in the algebraic sense. Moreover, the reverse path 


right to left. The area is 144 


( 


64.7 + 



28 


BASIC ENGINEERING THERMODYNAMICS 


p 


2-1 could not have been accompanied by the delivery of work by the 
system. 

The maximum-work process is a limiting process; in applying Eq. (2:3) 
to a constant-volume path, no value greater than zero may be assigned to 

i W 2 (or 2 W 1 ), as that notation applies 
in Fig. 2:2. On the other hand, no 
limit exists to the amount of paddle- 
wheel work which may accompany the 
tracing of the path in either direction, 
for heat flow outward may occur simul¬ 
taneously with work flow inward and 
increase the total amount of negative 
work required without limit. In at¬ 
tacking the problem of establishing the 
standards for the best possible per¬ 
formance of heat-work devices, our 
calculations are therefore based upon 
the maximum-work process as repre¬ 
senting the minimum wastage of the more costly form of energy. 

2:7. The specific heat at constant volume, c v , is sometimes nonrigor- 
ously defined as the amount of heat required to increase the temperature 
of unit weight of the substance by 1 deg while the volume remains con¬ 
stant. As has been demonstrated in the preceding article, that effect 
may be accomplished without heat flow or even with negative heat flow. 
A more rigorous definition would specify that the limiting, or maximum- 
work, process was to be followed as the heat was supplied. Under these 
conditions, dW = 0, and Eq. (2:5) becomes dQ = du. Thus a rigorous 
definition of the specific heat at constant volume would express it as 


Fig. 2:2. The constant-volume path 
CPdV = 0 ). 


Cy 


( du\ 
\dt ) v 


(2:7) 


The specific heat at constant volume is of principal importance when 
the substance is in the gaseous phase; it is not ordinarily expressed for 
liquids or solids. Since it has a definite value corresponding to each state 
of the system, it is a property of the system. 

Example 2:7. For the system and the process described in Example 1:10, calculate 
(a) the maximum work, ( b ) the maximum heat flow, and (c) the specific heat at con¬ 
stant volume, c v . 

Solution: 

(a) Since dV — 0, the maximum work is zero. [Eq. (2:6)] 

(b) Based on Eq. (2:3), since (ifF 2 )max = 0, 1 Q 2 = U 2 - U 1 = 30.6 Btu. [Ex. 1:10] 

10.2 


« * - (tr). - 


120 - 60 


= 0.17 Btu/(lb)(°F) or Btu/(lb)(°R) 






THE FIRST LAW AND THE CLOSED SYSTEM 


29 


2:8. Constant-pressure Processes. The area under the constant- 
pressure path, illustrated in Fig. 2:3, is P(V 2 - Vf) ovP 2 V 2 - PiV h and 
this is therefore the maximum work that may be developed. Substituting 
in Eq. (2:3), 

J iQz - (P 2 F 2 - P 1 V 1 ) = J(U 2 - Ui) 


1Q2 = u, 



-\u 1 + 


p 


f) 


The quantity L + PV/J is evidently a composite property since it is ex¬ 
pressed entirely in terms of properties. It is called enthalpy (en-thal'-py), 


and the symbol H will represent it. 
properties, enthalpy is also exten¬ 
sive; specific enthalpy, or enthalpy 
of unit weight, is denoted as h. 
Thus the following relations develop: 


Since U and V are both extensive 


p,>Pz 


H = U + 


PV 

J 


h — u -j- 


(iQOp, 


max work 


= H, 


H 1 


Pv 

T 

( 2 : 8 ) 

(2:9) 


An important warning must be in¬ 
serted here. Although examination 
of Eqs. (2:8) and (2:9) indicate that 
enthalpy is conventionally measured 


/ 

2 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 


V, V 2 V 

Fig. 2:3. The constant-pressure path 
[\PdV = P(V 2 - V x ) = P 2 V 2 - PxVx]. 


in Btu and that changes of enthalpy can be used to measure heat flow, 
under the special condition that a constant-pressure maximum-work process 
is followed, the reader must not fall into the error of thinking of the 
enthalpy as a kind of stored energy of the system. As a property, its 
values may be tabulated for each state of the system, and it finds useful¬ 
ness in ways such as may be inferred from Eq. (2:9); other uses of this 
property will develop later. 

Figure 2:4 illustrates schematically the manner in which the constant- 
pressure process may be conceived as a maximum-work process. As heat 
enters, the gas system confined below the weighted, frictionless piston 
expands slowly against the constant pressure created by the weight of the 
piston plus the pressure of the atmosphere acting on its upper surface. 
The maximum amount of work is performed by the expanding gas system 
and equals P(V 2 — V\); this is also the amount of work received by the 
external system that consists of the piston and the atmosphere. 

In the course of the process which was described in the preceding para¬ 
graph, heat has been received from some external system which at all 
times has a temperature higher than that of the confined system. If, 










30 


BASIC ENGINEERING THERMODYNAMICS 


/\i\Jeight\ 




after the confined system has completed its expansion to F 2 , a cold body 
replaces the hot body from which heat was received during the original 
process, the direction of flow of heat will be reversed; the confined system 

will again return, at constant pressure, to its original 
volume. As it does so, work is done on it by the 
piston and the atmosphere in the same amount 
that work was received by them during the expan¬ 
sion. Not only is the state path retraced but, as re¬ 
gards the enclosed gas system, the same is true of the 
process, for both the work and the net heat flow of 
the cycle are individually equal to zero. On the 
other hand, if all, or any part, of the energy received 
by the system during its expansion had been in the 
form of work, the system, though it could have re¬ 
turned to its original state over the same state path, 
could not have retraced the process; for the net work 
and net heat flow of the cycle would both be less 
than zero. 

2:9. The specific heat at constant pressure c 


Piston 




A® 


is 


Fig. 2:4. The con¬ 
stant-pressure maxi¬ 
mum-work process. 

the heat flow that will produce a temperature in¬ 
crease of one degree in one pound of the substance as a constant-pres¬ 
sure maximum-work process is carried out. Applying Eq. (2:9) in dif¬ 
ferential form to a unit weight of substance, 



( 2 : 10 ) 


The specific heat at constant pressure is a property that finds important 
uses in connection with all of the phases of the substance, solid, liquid, or 
gaseous. 


Example 2:9. The system described in Example 1:10 traverses a path at a constant 
pressure of 50 psig in the course of which its temperature changes from 60 to 120°F. 
Calculate (a) the final volume of the system, (6) the maximum work that could 
accompany the process, (c) the maximum heat flow during the process, ( d ) the change 
of enthalpy of the system, (e) the specific heat at constant pressure for this substance. 
(/) Write an equation of state for this substance that will express the enthalpy in 
terms of the pressure and the volume. 

Solution: 


(a) v 2 = 


CT 2 
V 2 


0.37(120 -j- 460) 
50 + 14.7 


= 3.31 ft 3 /lb; V 2 = 9.93 ft 3 


(b) (iW 2 )nuLx = P(V 2 - Vi) = (64.7)(144)(9.93 - 8.91) = 9500~ft-lb 

(c) u 2 = A + Bp 2 v 2 = 16 + (0.46) (64.7) (3.31) = 114.7 Btu/lb 

ut-ui = 114.7 - 104.5 = 10.2 Btu/lb; U 2 - XJ x = (3)(10.2) = 30.6 Btu 















THE FIRST LAW AND THE CLOSED SYSTEM 


31 


From Eq. 2:3, 

(lQ2)max = f 2 — U l + (lTF 2 )max/«/ = 30.6 + "^VlT = 42.8 BtU 

(d) From Eq. (2:9), H 2 - H x = ( 1 Q 2 ) Pi max = 42.8 Btu. 

, * /A h\ 42.8 

(e) - (a t) P - 8 (120 - 60) - °- 238 B ‘u/(lb)CF) 

(/) h = u + Pv/J = A + Bpv + Pv/J = 16 + 0.46py + 144pv/778 = 16 + 0.645/w 

2.10. The Constant-temperature (Isothermal) Process. The system 
can be caused to remain at constant temperature if the cylinder in which it 
is confined is immersed in a constant-temperature bath and if the cylinder 
walls are perfect conductors of heat; i.e., only an infinitesimal difference 
in temperature between system and bath 
is necessary to cause heat flow at any 
desired rate. The path traversed dur¬ 
ing a constant-temperature process may 
be illustrated as in Fig. 2:5, but it will be 
observed that this method of plotting 
does not lend itself to the direct calcula¬ 
tion of jP dV and thus enable the calcu¬ 
lation of the maximum work that can ac¬ 
company the performance of the path. 

However, for a simple system, all prop¬ 
erties are established if any two that are 
independent of each other are known and 
to each point along the TV path a defi¬ 
nite pressure is therefore attached. Assuming that an equation of state 
such as Eq. (1:6), connecting the temperature, pressure, and volume of 
the substance of which the system is composed, is available, these pres¬ 
sures can be determined and an equivalent PV path laid out. From this 
path the maximum work may be calculated. The change of internal 
energy is dependent only on the initial and final states and may be deter¬ 
mined when a suitable equation of state [as Eq. (1:7), for example] is 
known to apply to the substance. Substitution in Eq. (2:3) will then 
make possible the calculation of the maximum heat flow that can accom¬ 
pany the process. The introduction of paddle-wheel work into the proc¬ 
ess will not change AU if the end states are the same as for the maximum- 
work process but will reduce the work and the heat flow by amounts 
corresponding to the amount of such paddle-wheel work. 

Example 2:10. Assume a system composed of 2 lb of a substance for which the 
equations of state are represented by Eqs. (1:6) and (1:7). Let it expand at a con¬ 
stant temperature of 50°F (isothermally) from an initial pressure of 100 psia and 
volume of 4 ft 3 to a final volume of 8 ft 3 . Determine (a) the equation of the path in 
terms of p and v, ( b ) the final pressure of the system, (c) the change of internal energy 


T 


Fig. 2:5. The 
ature path. 


1/ 

constant-temper- 






32 


BASIC ENGINEERING THERMODYNAMICS 


that takes place, and ( d ) the maximum work and maximum heat flow which can 
accompany the process. 

Solution: 

(а) For the purpose of substituting in Eq. (1:6), pi = 100; v x — \ = 2; T\ = 50 +460 
= 510 

C = pv/T = (100)(2)/510 = 0.393 

The equation of the path is pv = 200 or pV = 400 and PY = (144)(400) = 57,600. 

(б) Since piVi = p 2 v 2 , p 2 = p\V\/v 2 = (100)(2)/4 = 50 psia. 

(c) Applying Eq. (1:7), u\ — A + B(p x v i), and u 2 = + + B(p 2 v 2 ). But, for this 
substance , piVi = p 2 v 2 during an isothermal, and A u is therefore zero. [The reader is 
warned that this result is obtained only because of the special form of Eqs. (1:6) and 
(1:7) and it will not generally follow that an isothermal path is also one which is traced 
through states having the same internal energy.] 

(d) (iJF 2 )max = [ V2 PdV = 57,600 \ V * dV/V = 57,600 log, f = 39,900 ft-lb and, 

JV i JV i 

applying Eq. (2:3), 

( 1 Q 2 W = U + (iTF 2 )max// = 0 + 39,900/778 = 51.3 Btu 

2:11. Constant-internal-energy Processes. The closed system can 
traverse a constant-internal-energy path only by maintaining an exact 
balance in the rate at which heat and work cross its boundaries; i.e. } the 
rate of heat flow, if inward, must be exactly equal to the rate at which 
work is delivered by the system to other systems. Thus, applying 
Eq. (2:3), J 1 Q 2 = iW 2 . The path may be illustrated as in Fig. 2:6. 

u 


/ 2 


V 

Fig. 2:6. The constant-internal-en¬ 
ergy path. 



Fig. 2:7. The zero-work constant-inter¬ 
nal-energy process. 


The calculation of the amount of work that can accompany the tracing 
of the constant-internal-energy path requires that the path shown in Fig. 
2:6 be transferred to pressure-volume coordinates. This can be done if an 
equation of state such as Eq. (1:7) is available; JP dV can then be evalu¬ 
ated. The maximum heat flow will be equal to the maximum work when 
expressed in terms of the same energy unit. 
































THE FIRST LAW AND THE CLOSED SYSTEM 


33 


A special type of constant-internal-energy process is of interest. Sup¬ 
pose the system originally to be confined in a section of a container with 
rigid nonconducting walls. This vessel is designated as A in Fig. 2:7 and 
is connected with a vessel B, also with rigid nonconducting walls, through 
an insulated passage that includes a closed valve. Originally, vessel B 
has been voided of all matter. The valve is now opened, and the mole¬ 
cules of the system will rush in to occupy this voided space until the pres¬ 
sure in the two vessels is equalized. The resistance of no external system 
has been overcome, and thus the amount of work involved is zero. The 
reader may find it helpful to conceive of work as done on an imaginary 
piston which is moving with the speed of the molecules themselves and 
therefore offers no resistance to the expansion of the system. Since the 
passage of heat was prevented by the nonconducting walls of both vessels, 
the internal energy, according to Eq. (2:3), must have remained constant. 
For this special case, not only A U but also iQ 2 and iW 2 are zero; yet the 
state of the system has changed. 

The path followed during this process would have been difficult to 
trace, but points along that path could 
be located by allowing the total ex¬ 
pansion to take place by stages. 

That is, a series of partitions could be 
provided in the voided chamber, and 
the expansion of the system could be 
halted at each successive partition 
long enough to allow an equilibrium 
state to be reached. The pressure 
and the volume could then be meas¬ 
ured and a point on the pressure-vol¬ 
ume graph plotted. A path drawn 
through these points would agree ex¬ 
actly with that followed during a maximum-work process, but, of course, 
the area under it would not have the same significance. 

2:12. The Adiabatic Process. By definition, the adiabatic process 
involves no heat flow. Thus, from Eq. (2:5), dU = -dW/J. If this is 
a maximum-work process, P dV may be substituted for dW, and it 
follows that 

JdU= -PdV 



Fig. 2:8. The maximum-work adia¬ 
batic process. 


or 


dU _ _ P 

dV J 


( 2 : 11 ) 


Thus, for the maximum-work adiabatic process, the path will have the 
form shown in Fig. 2:8, the slope at any point being negative and propor¬ 
tional to the pressure. Assuming that a suitable equation of state is 





34 


BASIC ENGINEERING THERMODYNAMICS 


available, a corresponding path may be traced on a graph having pressure- 
volume coordinates, and the maximum work that may accompany the 
process can be determined. 

When an algebraic equation of state such as Eq. (1:7) is available, 
du/dv may be determined from it as it applies to any process and, for the 
special case of the maximum-work adiabatic, placed equal to —P/J. 
From this equality the equation relating pressure and volume during a 
maximum-work adiabatic may be obtained. 


Example 2:12 A. Using Eq. (1:7), derive the equation that applies to the maximum- 
work adiabatic process when the system is composed of an ideal gas. 

Solution: 


u = A -f- Bpv 

du = Bp dv -f Bv dp (any process) 

^ = B (^p + v — ^ = — j (maximum-work adiabatic process) 


or 


B { V + V t) - -#§P = 0.185p 

and, dividing by p, 

b( i + ?$0- -0-185 

\ p dv / 


from which 


B + 0.185 = _ vdp 
B p dv 

Dividing by pv and transposing, 

dp B + 0.185 dv _ 

~p B 7 ~ 

Integrating, 

, , B + 0.185 , 

loge p H- -g - loge V = const 

or 

pv k = const 
when 


B + 0.185 
B 


[Eq. (1:7)] 
[Eq. (2:11)] 


( 2 : 12 ) 


(2:13) 


The path shown in Fig. 2:8 applies to the maximum-work adiabatic 
only. The process which it represents may be traversed in either direc¬ 
tion. During the expansion 1-2, work is done by the system at the 
expense of its internal energy; during the compression 2-1, the increase of 
internal energy exactly equals the work supplied by external systems. 

Example 2:12 B. The system described in Example 1:10 has an initial pressure of 
50 psig and an initial temperature of 60°F as in that example. As the result of a 







( 


THE FIRST LAW AND THE CLOSED SYSTEM 


35 


maximum-work adiabatic process, its pressure decreases to atmospheric pressure. 
Calculate (a) the final volume of the system, (6) its final temperature, (c) the work that 
accompanies the process, ( d ) the change of internal energy of the system as a result 
of the process. 

Solution: 


(a) From Eq. (2:12), k = (0.46 + 0.185)/0.46 = 1.4. 

, , / mV / 1 - 4 /64 7\ 0 - 713 

Pivi 1 - 4 = V& 2 1A or v 2 = v x = 2.97 ( j~) = (2.97)(2.88) = 8.55 ft 3 /lb 

V 2 = (3) (8.55) = 25.65 ft 3 


(b) T 2 = V 2 V 2 /C = (14.7)(8.55)/0.37 = 339°R or -121°F 

f Vi 

(c) The process takes place at maximum work, and therefore \W 2 — / P dV. 

JV 1 

pv k = const = Pl vi* = (64.7) (2.97) 14 = (64.7) (4.6) = 297 
f = 144p = ^?7) = 42^00 F = 

V IA V \A > 

r v 2 fv 2 dr) r 8.55 du 

lWi - J Vt PdV = (42,700) (3) l ^ = 128,100 

“ (128,100K-2.5) ( g -A_ 

= 71,700 ft-lb = 92 Btu 


2.97° 4 / 


(d) U*-Ui = 3(u 2 - wx) = 3B(p 2 y 2 - pivi) = (3)(0.46)[(14.7)(8.55) - (64.7)(2.97)] 
= -92 Btu 


If the work accompanying an adiabatic is less than the maximum, a 
multitude of paths may be traced by the system but these paths will not 
be retraceable as adiabatic processes. For example, a system may be 
increased in temperature at constant volume adiabatically if the paddle- 
wheel process is used to supply the necessary energy; the reader’s own 
experience can hardly fail to tell him that, although the path may be 
retraced, it cannot be retraced as an adiabatic. Another example is the 
unrestrained and adiabatic expansion at constant internal energy described 
in Art. 2:11; after the system has expanded to occupy the full volume of 
both vessels, it cannot be restored to its original pressure and volume at 
constant internal energy by any process which does not involve work and 
heat flow. 

2:13. The Paddle-wheel Process and Friction. The paddle-wheel 

process has been used in preceding pages to illustrate one of the two rea¬ 
sons why the work with respect to a system as a given state path is traced 
may be less than the maximum; the other was found in the difference 
in the pressure that the system exerted against its moving and its station¬ 
ary boundaries as during an expansion which was wholly or partially 
unrestrained. 

The paddle-wheel process as described heretofore was a calculated 
process in which the effects of intermolecular friction were deliberately 
invited to accomplish a certain effect. That effect was to bring about a 
change in the state of the system, by supplying work to it, which might 









36 


BASIC ENGINEERING THERMODYNAMICS 


have been produced by heat flow alone. Since this has been condemned 
as wasteful (energy being more valuable in the form of work than as 
heat), some confusion may be caused in the mind of the reader as to why 
the paddle-wheel process was ever contemplated. The explanation is 
that the paddle-wheel process simply exemplifies the effects of friction 
wherever and however it may develop. Friction is always present in 
practical apparatus in some degree although it may be reduced toward 
zero as a limit as, for instance, by improved methods of lubrication. 

As the piston of Fig. 2:1 moves, during either an expansion or a com¬ 
pression of the system, some friction will develop between it and the 
cylinder walls. The effect of that friction is the same as that of a paddle 
wheel since it reduces the net work of the process (in the algebraic sense) 
regardless of the direction of piston movement and because the equivalent 
amount of energy is changed to the form of heat. This is a wasteful 
process, as we have observed, but at least it has the virtue of being 
unavoidable rather than deliberate. The paddle-wheel process thus 
represents the effects of friction as they bring about deductions from the 
maximum work that can accompany a given path, and we may revise an 
earlier statement to the form: Friction and unrestrained expansion will 
account for a reduction in the work that accompanies the tracing of a given 
path from the maximum , or jP dV. 

2:14. The Closed-system Cycle. The power plant has the function of 
delivering work continuously. Power is the word that is used to describe 

the rate at which work is delivered by 
such devices. An important power 
unit is the horsepower, which is arbi¬ 
trarily defined as a rate of work de¬ 
livery equivalent to the delivery of 
550 ft-lb each second. The kilowatt, 
equivalent to 1.34 hp, is also fre¬ 
quently used. The horsepower hour 
is the total amount of energy repre¬ 
sented by the delivery of work at the 
rate of 550 ft-lb/sec (1 hp) over a time 
period of 1 hr, or 3600 sec. A similar 
unit, 1.34 times as large, is the kilo- 
watthour; this is the unit in terms of which energy is usually bought and 
sold. It will be noted that it is energy that is purchased although the 
rate at which that energy is accepted by the consumer is often a factor 
in establishing the unit price per kilowatthour. 

In order to deliver work continuously, the power plant operates on some 
sort of cycle, returning the system periodically to its original state. One 
closed-system cycle is illustrated in Fig. 2:9, the cycle being composed of 



Fig. 2:9. A closed-system cycle. 









THE FIRST LAW AND THE CLOSED SYSTEM 


37 


two constant-pressure and two constant-volume paths. Assuming that 
the processes are carried out with maximum work, which would be possi¬ 
ble in the actual power plant only as a limiting condition, the rate of work 
delivery per cycle is represented by the area enclosed within the cycle, for 
the (positive) work during the constant-pressure expansion 1-2 is the area 
a-1-2-6 and the (negative) work accompanying the constant-pressure 
compression 3-4 is area 6-3-4-a; no work enters into the constant-volume 
processes 4-1 and 2-3. 

From the First Law, the area enclosed within the cycle also represents 
the net heat flow (heat flow in minus heat flow out) during the cycle. 
When the processes that comprise the cycle of Fig. 2:9 are analyzed on 
the basis of the First Law (and as maximum-work processes), it will be 
found that heat entered the closed system during processes 4-1 and 1-2 
and left the system during processes 2-3 and 3-4. This implies that a 
change in the surroundings took place in the course of the cycle, for the 
temperature of those surroundings must have been higher than the tem¬ 
perature of the system during processes 4-1 and 1-2, to account for the 
entry of heat, and lower than the temperature of the system during 
processes 2-3 and 3-4, as heat left the system. It may be shown that the 
highest temperature reached by the system in the course of the cycle is at 
point 2. If an external reservoir is available from which heat may be 
withdrawn at or above this temperature it will be capable of supplying 
the heat necessary to carry out processes 4-1 and 1-2; to this external 
reservoir is given the name of the source. Similarly, the lowest tempera¬ 
ture reached by the system during the cycle is that corresponding to point 
4, and the heat that leaves the system during processes 2-3 and 3-4 could 
be transferred to a reservoir at or below this temperature; this reservoir 
is termed the sink, or refrigerator. It will later be shown that at least one 
source and one refrigerator, which must differ in temperature, are 
required for any power cycle. It will follow that during any such cycle 
there must be a two-way flow of heat; that, during a certain process or 
processes of the cycle, heat must be accepted from the source and, during 
another part of the cycle, heat must be rejected to the refrigera¬ 
tor. Further, since for the power cycle the net flow of heat must be posi¬ 
tive, the heat received from the source must exceed that rejected to the 
refrigerator. 

2:15. The heat engine is defined as a device that has as its purpose the 
continuous production of work from heat and across the boundaries of 
which only heat and work may pass. In Fig. 2:10 the apparatus required 
for a heat engine that operates on a closed-system cycle is illustrated. 
The cylinder walls are nonconducting, but the head permits the passage 
of heat. A nonconducting cap is provided for the head so that heat flow 
can be entirely cut off if one or more of the processes traversed in the 


38 


BASIC ENGINEERING THERMODYNAMICS 


course of the cycle is an adiabatic. Provision is made for rapidly shifting 
the source S into place at the head of the cylinder when the cycle requires 
that heat enter the system. Similarly, the refrigerator R is placed in that 

position when heat is to be re¬ 
moved. The limits of piston mo¬ 
tion are as shown by the full and 
the dashed lines of the sketch. 

It is evident that, while the 
closed-system cycle fulfills the tech¬ 
nical requirements for a heat en¬ 
gine, it has serious practical limi¬ 
tations since the chamber in which 
work is performed must also act as 
a heat exchanger. Only a few 
actual engines are based on the 
closed-system cycle, and they are 
of small importance from the stand¬ 
point of practical performance. 
However, from the speculative standpoint, we may profit greatly from a 
study of the limits of performance set by theory for the operation of this 
type of heat engine. 

2:16. Efficiency of a Heat Engine. The performance of heat engines is 
compared in terms of their efficiencies. The efficiency 77 of a heat engine is 
defined as the rate at which it delivers work divided by the rate at which 
it receives heat from the source, or 

W 

v = JQs (2:14) 




s 


R 



Fig. 2:10. The closed-system heat en¬ 
gine. 


in which W is the rate of work delivery, in foot-pounds per cycle or per 
unit of time, and Qs is the rate at which heat is supplied the engine from 
the source, in Btu per cycle or per unit of time. 

Since, by the First Law, J J'dQ = ,fdW for any cycle, the work of the 
cycle is the difference between the rate of heat supply from the source, 
JQs, and the rate of heat rejection to the refrigerator, JQr, and an alter¬ 
nate expression for the efficiency is 


V = 


Qs Qr 
Q s 


(2:15) 


Either Eq. (2:14) or Eq. (2:15) may be used to calculate the efficiency 
of any heat engine according to the relative convenience of their applica¬ 
tion to that purpose. 

Example 2:16. A heat engine causes the system described in Example 1:10 to 
traverse clockwise a cycle similar to that illustrated in Fig. 2:9. The engine makes 























THE FIRST LAW AND THE CLOSED SYSTEM 39 

200 rpm, and Vl = F 4 = 37 ft*, V 2 = F 3 = 60 ft*, Pl = p 2 = 30 psia, p 3 = 

15 psia. Assume all processes to be maximum-work processes, and calculate 
(a) the net work of the cycle, ( b ) the horsepower output of the engine, (c) the tempera¬ 
tures at points 1, 2, 3, and 4 around the cycle, ( d ) the heat supplied per cycle, Qs, ( e) 
the rate of heat rejection per cycle, Qr, (/) the efficiency of the cycle. 

Solution: 


(a) i\V i = 2 W 3 = 0 (maximum-work constant-volume processes) 
iW 2 = P X {V 2 - Vi) = (30)(144)(60.- 37) = 99,400 ft-lb 
3 Wi = P 3 (F 4 - Vi) = (15) (144)(37 - 60) = -49,700 ft-lb 


Net work of cycle 
(49,700) (200) 

(550) (60) 30 1 hp 

(c) T x = = 1000°R, or 540°F 

T 2 = = 1620°R, or 1160°F 

rji _ (15) (60) _ Qin°T> ornori 

Tz (0.37) (3) _ 810 R ’ ° r 350 F 

rp _ (15) (3/) _ rfinoT) /jnori 

4 (0.37) (3) ~~ 500 R ’ ° r 40 1 


= 49,700 ft-lb = 64 Btu 


id) Qs = aQi + 1 Q 2 = Ui - U 4 + H 2 - Hi = Mc v (Ti - T 4 ) + Mc p (T 2 - Ti) 

= (3) (0.17) (1000 - 500) + (3) (0.238) (1620 - 1000) = 698 Btu 
(e) -Qr = 2 Q 3 + sQi = U z - U 2 + H, - H 3 = Mc v (T 3 - T 2 ) + Mc v {T A - T s ) 
= (3)(0.17)(810 - 1620) + (3)(0.238)(500 - 810) = -634 Btu 
Qr = 634 Btu 


Qr might also have been found as the difference between Q s and the net work of the 
cycle, or Q R = 698 — 64 = 634 Btu. 

(/) Efficiency of the cycle = = 698 634 = 0.0917, or 9.17 per cent. 

Qs 698 

Problems 

1. Referring to Prob. 9, Chap. 1, parts a , b, c, e, f, i, j, and k, does E increase, de¬ 
crease, or remain constant in each case? 

2. Figure 1:2 describes a state path graphically. Which of the following are suffi¬ 
cient to convert it to a description of the closed-system process as well? (a) The heat 
flow iQ 2 is known. ( b ) The work AV 2 is known, (c) Both 1 Q 2 and iW 2 are known. 
( d ) Neither is known. 

3. At the beginning of a process, the volume of a closed system is 10 ft*, and the 
pressure is 200 psia. At the end, V 2 = 15 ft*, p 2 = 200 psia. The net work that 
accompanies the process is +100,000 ft-lb. Calculate the heat flow during the process 
if the system consists of (a) 16 lb of gas IF; (6) 7 lb of gas X ; (c) 11 lb of gas Y; ( d ) 
1.5 lb of gas Z. 

4. During a certain process, 150,000 ft-lb of work enters a closed gas system, and 
50 Btu of heat leaves it. What change takes place in the temperature of the system 
and in its internal energy if the system consists of (a) 10 lb of gas W; (6) 4 lb of gas X; 
( c ) 8 lb of gas Y ; (d) 2 lb of gas Z ? 

5. A closed system traverses a cycle which is composed of four processes. The 
first process is an expansion during which 100 Btu of heat enters the system and 25,000 
ft-lb of work leaves the system; the second process, also an expansion, is accompanied 










40 


BASIC ENGINEERING THERMODYNAMICS 


by no heat flow while 30,000 ft-lb of work leaves the system; the third, a compression, 
sees heat leaving the system in the amount of 80 Btu while 35,000 ft-lb of work enters 
across the boundary; the fourth process is adiabatic. What amount of work accom¬ 
panies the fourth process? 

6. The temperature of a 1-lb closed system increases by 150°F during a process 
which is accompanied by an outward heat flow of 35 Btu. Calculate the work of the 
process if the system consists of (a) gas W ; (6) gas X; (c) gas Y; (d) gas Z. 

7. The temperature of a closed gas system weighing 2 lb decreases by 100°F during 
a process. What is the difference between tfle heat flow and the work of the process if 
the system consists of (a) gas W ; (6) gas X; (c) gas Y ; (d) gas Z? 

8. In Prob. 7, if the process takes place at constant volume, what maximum heat 
flow may accompany it in each case? 

9. A closed container with rigid nonconducting walls contains a system consisting 
of a gas and an electrical heating element. Current is supplied the latter from a source 
outside the container; as a result, the temperature and pressure of the gas rise. Is 
the work of the process a maximum? Is the heat flow a maximum? Is the equiva¬ 
lent of paddle-wheel work performed? If so, what special kind of friction is involved? 
How does the system differ from that of Art. 2:3? 

10. A closed system expands at a constant pressure of 120 psia from an initial 
volume of 1.0 ft 3 to a final volume of 1.4 ft 3 . What maximum work could accompany 
the process? If the system is returned to its original state over the same state path, 
what is the maximum work for the second process? What is the maximum work of 
the cycle formed by the two processes? 

11. At a constant pressure of 1 atm, the volume of a closed system changes from 20 
to 17 ft 3 . What maximum work may accompany the process? 

12. A closed system changes from an initial pressure of p i = 40 psia and volume 
Vi = 5 ft 3 to a final pressure p 2 = 20 psia and volume V 2 = 10 ft 3 over a state path 
which is a straight line on pV coordinates. What maximum amount of work may 
accompany the process? 

13. The end states are the same as in Prob. 12, but the path is hyperbolic with 
pV = const. Calculate the maximum work. 

14. The initial and final states of an expanding closed system are as in Prob. 10, 
but the path is a circular arc, convex upward and having its center at p c — 120 psia, 
V c = 1.2 ft 3 . What maximum work may accompany the process? If the system is 
returned to its initial state over a circular arc having the same center but concave 
upward, what maximum work may accompany this second process? What maximum 
work may accompany a cycle formed of the two processes? 

15. If the expansion of Prob. 14 takes place in a cylinder fitted with a piston which is 
backed by the pressure of the atmosphere (14.7 psia), what maximum useful work 
could accompany the expansion? On the return process, what is the maximum useful 
work? What is the maximum useful work of the cycle? Compare with the work of 
the cycle as calculated in Prob. 14. 

16. A part of the resistance offered by the piston to the expansion of the system of 
Prob. 10 is due to the pressure of the atmosphere (14.7 psia) on its back face. What 
maximum useful work could accompany the expansion? What maximum useful 
work could accompany the return process? What is the maximum useful work of the 
cycle? Compare with the work of the cycle as calculated in Prob. 10. 

17. During a process that takes place at constant volume, the temperature of a 
closed system increases by 100°F. What maximum heat flow may accompany the 
process if the system consists of (a) 5 lb of gas W] (b) 3 lb of gas X; (c) 4 lb of gas Y ; 
(d) 1 lb of gas Z1 


THE FIRST LAW AND THE CLOSED SYSTEM 41 


18. Calculate the specific heat at constant volume, c v , of (a) gas W; ( b ) gas X; (c) 
gas Z. 


19. If v represents the rate at which heat flows across the boundaries of a 1-lb 
closed system during a constant-volume process, which of the following relations 


are not possible? (a) (§\ > *; (i» (f \ < «.; (c) (f \ = c; W (f),. < 0. 

20. During a process that takes place at constant pressure, the temperature of a 
closed system increases by 100°F. What maximum heat flow may accompany the 
process if the system consists of (a) 5 lb of gas W ; (b) 3 lb of gas X; (c) 4 lb of gas Y ; 
(d) 1 lb of gas Z? 

21. Write an equation that will express the specific enthalpy as a function of the 
pressure and the specific volume for (a) gas W; (6) gas X ; (c) gas Z. Calculate the 
specific heat at constant pressure for each of these gases. 

22. A closed gas system changes from p x = 150 psia, V\ = 2 ft 3 , to p 2 = 100 psia, 
F 2 = 1 ft 3 . Calculate the change of enthalpy of the system if it is composed of (a) 
gas W] ( b ) gas X; ( c ) gas Y; ( d ) gas Z. 

23. The isothermal expansion of a closed gas system starts from pi = 200 psia, 
V\ = 1 ft 3 , and ends when the volume becomes 10 ft 3 . What are the maximum work 
of the process, the maximum heat flow, and the change of internal energy of the sys¬ 
tem, if it is composed of (a) gas W; (b ) gas X; (c) gas 7; (d) gas Z? 

24. A closed gas system is compressed isothermally at 200°F from pi = 15 psia to 
Pi = 75 psia. Calculate the maximum work and the maximum heat flow of the 
process and the change of enthalpy of the system if the system is composed of 1 lb 
of (a) gas W; (6) gas X; (c) gas F; (d) gas Z. 

25. Show that the maximum work of an isothermal process is given by 


iW 2 = 144 MCT log, ^ 

P2 

if the system is composed of an ideal gas. 

26. A simple closed system which is not composed of an ideal gas undergoes an 
isothermal process. Which of the following statements are necessarily true? (a) 
J 1 Q 2 = 1 W 2 ; (b) AU = 0; (c) T 2 = 7\; (d) Vl Vi = p 2 V 2 . 

27. A simple closed system not composed of an ideal gas goes through a process 
during which its internal energy remains constant. Which of the following statements 
are necessarily true? (a) JiQ 2 = \W 2 ; ( b ) U 2 = lh; (c) T 2 — 7\; ( d ) p x V 1 = p 2 V 2 . 

28. Does Eq. (2:11) apply to a closed system which is not composed of an ideal 
gas? Is it necessary that it be a simple system? Is the storage of energy due to 
motion, gravity, capillarity, magnetism, or electricity prohibited when this equation is 
to be applied? What changes, if any, would you suggest if it is to apply to (a) fluids 
in general; ( b ) complex systems; (c) systems which may store energy due to motion, 
etc.? 

29. What is the value of the exponent k in the maximum-work adiabatic process if 
the closed system is composed of (a) gas IF; (6) gas X; (c) gas Z? 

30. Show that, for an ideal gas, the maximum work that may accompany a closed- 
system adiabatic process may be expressed as 

w 144( Pl Fi - p 2 V 2 ) 144MC(Ti - T 2 ) 
lW * ~ k - 1 = k - 1 

31. A closed gas system has an initial pressure of 100 psia, and an initial volume of 
1 ft 3 . It undergoes a maximum-work adiabatic process, reaching a pressure of 20 psia. 




42 


BASIC ENGINEERING THERMODYNAMICS 


Calculate the final volume, the work of the process, and the change of internal energy 
of the system if composed of (a) gas W; (b ) gas X; ( c ) gas Y ; ( d ) gas Z. 

32. A closed system goes through an adiabatic process. Which of the following 
statements is necessarily true? (a) X W 2 = J(Ui — U 2 ); ( b) pV k = const; (c) 1 Q 2 = 0. 
Explain your answers. 

33. A simple closed system expands adiabatically and at constant atmospheric 
pressure. The initial volume is 7 ft 3 , the final 10 ft 3 . Calculate the work of the 
process and the change of internal energy if the system is composed of (a) gas W ; 
( b ) gas X;(c) gas Y; ( d) gas Z. 

34. A simple closed gas system undergoes an adiabatic process during which a con¬ 
stant-volume state path is followed. During the process the temperature increases 
by 100°F. Calculate the work of the process if the system weighs 1 lb and is composed 
of (a) gas W ; ( b) gas X ; (c) gas Y ; ( d) gas Z. 

35. For an ideal gas, express the ratio of the final to the initial absolute tempera¬ 
ture, T 2 /T 1 , as the result of a maximum-work adiabatic process as (a) a function of 
the pressure ratio, p 2 /pi ; ( b) a function of the volume ratio, V 2 /V x . 

36. Prove that for an ideal gas the value of k equals the ratio c p /c v . 

37. What is the energy equivalent, in Btu, of the horsepower-hour? Of the kilo- 
watthour? 

38. A heat engine operates on a cycle in the course of which it delivers 50,000 ft-lb 
of work to external systems during certain of the processes which make up the cycle 
and receives 30,000 ft-lb of work from the surroundings during the other processes of 
the cycle. The efficiency of the cycle is 15 per cent, and it is traversed 150 times per 
minute. What amount of heat is received from external systems in the course of the 
cycle? What is the net heat exchange with external systems during one complete 
cycle? What horsepower does the engine deliver? 

39. A heat-engine cycle is composed of four closed-system processes, 1-2, 2-3, 3-4, 
and 4-1. If 1 Q 2 = 75 Btu, 2 Q 3 = —5 Btu, 3 Q 4 = —48 Btu, and 4 Qi = —2 Btu and 
the cycle is traversed three times per second, calculate its efficiency and the horsepower 
output. 

40. What is the efficiency of the cycle described in Prob. 5? 

41. A heat-engine cycle is composed of three maximum-work closed-system proc¬ 
esses. An isothermal expansion begins at p x = 100 psia, V x = 1 ft 3 , t\ = 1500°F, 
and ends at p 2 = 20 psia. The next, a constant-pressure process, ends at V 3 = 1 ft 3 . 
The process that closes the cycle takes place at constant volume. If the system is 
composed of (a) gas W, ( b ) gas X, ( c ) gas Y, ( d ) gas Z, calculate the work and the 
heat flow accompanying each process, the net work of the cycle, and its efficiency. 
If a single source and a single refrigerator, each at respectively constant temperature, 
are employed, what minimum temperature is possible for the former, what maximum 
temperature for the latter? 

42. In Prob. 41 consider that, owing to friction, the amount of work delivered during 
the isothermal expansion is reduced by 10 per cent and the amount of work received 
from external systems during the constant-pressure compression is increased in the 
same ratio. The constant-volume process is still a maximum-work process. How 
does this change affect the efficiency of the cycle when each of the four gases consti¬ 
tutes the working substance? 

Symbols 

A area; also, a constant 

B a constant 

c p specific heat at constant pressure 

c v specific heat at constant volume 


THE FIRST LAW AND THE CLOSED SYSTEM 


43 


C a constant 

e stored energy of a system of unit mass, in general (thermal units) 

E stored energy of a system, in general (thermal units) 
h enthalpy of a system of unit mass (thermal units) 

H enthalpy of a system (thermal units) 

J proportionality factor 
k a constant ratio 
p pressure, psi 

P pressure in general; also, in psf 
Q heat flow (thermal units) 

Qs rate of heat flow, from source (thermal units) 

Qr rate of heat flow, to refrigerator (thermal units) 
t scalar temperature 
T absolute temperature 

u energy stored as thermal energy in unit mass; specific internal energy (thermal 
units) 

U energy stored as thermal energy in a system; internal energy (thermal units) 
v volume of unit mass; specific volume 
V volume of a system 
W work; also, rate of work delivery 

Greek letters 

77 efficiency of a heat engine 

Subscripts 

max maximum 

p constant pressure 
v constant volume 


CHAPTER 3 


THE FIRST LAW AND THE OPEN SYSTEM 


3:1. Stored Energy of the Open System. The open system has been 
described in a manner such as to permit the flow of matter as well as of 
heat and work across its boundaries. For that reason the open-system 
process is called a flow process, while the process which involves a closed 
system is termed a nonflow process. From the use of the word flow it 
may be inferred that motion may no longer be ignored as in the case of the 
simple closed system and that account must be taken of energy stored in 
the system by reason of that motion. The amount of this form of stored 
energy may be measured in terms of the work required to start the system 
from rest and give it the designated velocity. Thus velocity is an addi¬ 
tional property concerning which information must be at hand in dealing 
with an open system. It is not, however, one of the properties included 
in any equation of state of the pure substance and therefore cannot be 
established from the values of other properties alone; it must be individu¬ 
ally stated or calculable. 

In calculating the work required to impart motion to a body (and 
thereby the energy stored in the body by reason of its motion) we return 
to the laws of motion and find that 

F = ma 


where, if m is the mass in slugs and a is the acceleration in feet per second 
per second, F is the force required to impart that acceleration in pounds. 
If the time interval is short, the acceleration is constant and equal to 
dV/dt (the instantaneous change of velocity with respect to time), dl is 
the distance traversed, and V = dl/dt. But the work delivered to the 
body during this time interval is 


and 


Thus 


or 


F dl = ma dl 


dV = dVdl = dV y = VdV 
dt dl dt dl dl 


dW = F dl = mV 



dl = mV dV 



W = 


mV 2 


A E 


2 


44 








THE FIRST LAW AND THE OPEN SYSTEM 


45 


For a unit-weight (1-lb) system the energy stored by reason of its motion 
is therefore V 2 /2g. 

Although gravity may usually be neglected, it is occasionally necessary 
to consider its effects. In this case the energy stored is the work required 
to elevate the system from the datum level, or the product of its weight 
and the change in elevation. In the case of unit weight the energy storage 
is simply the elevation, denoted by z. 

Thus, for the slightly less simple system for which motion and gravity 
are taken into account (although capillarity, magnetism, and electricity 
are still ignored), 

e = W + 27 g + J (3:1) 

3:2. Work and the Open System. Work may cross the boundaries of 
the open system owing to a movement of those boundaries (an expansion 
or a contraction of those boundaries) as in the case of the closed system. 

The flow of work across the boundary may also be due to paddle-wheel 
action as in the case of the closed system, but, for the open system, work 
may not only enter but also leave the system by this method under certain 
conditions and subject to certain limitations. For if the flow is continu¬ 
ous and follows a definite pattern throughout the system, it may be used 
to do work on the paddles and the system may in this manner deliver 
work to external systems. 

The fluid that enters across the boundaries of the open system does so 
against the resistance offered by the pressure of the system at the point of 
entry; this requires that work be done on the system to force this entry. 
The system itself does work in 
ejecting fluid which crosses its 
boundaries in an outward direction. 

This type of work is called flow 
work. 

For the purpose of convenient 
reference, the work of the open 
system exclusive of flow work will 
be called external work , and the 
symbol W will be used to represent it; external work may include work 
flow produced by either or both of the methods discussed in the first two 
paragraphs of this article. 

3:3. Flow Work. In Fig. 3:1 is shown an elementary weight dM of 
fluid which is about to enter a large system S across its boundaries B. 
The flow work may be calculated as that which the piston shown in the 
sketch must perform to force that entry. Because of the relative size of 
S and dM, the pressure of the large system will not change appreciably as 







46 


BASIC ENGINEERING THERMODYNAMICS 


the element of fluid is introduced. Since the boundary is assumed 
entirely permeable, it will not offer resistance of its own to the passage of 
dM and the piston will therefore operate against the constant pressure P 
(the local pressure of the large system at the point of entry). The volume 
swept through by the piston is v dM, and thus the work it performs is 
Pv dM. For the passage of a finite quantity of fluid, 

Flow work = fPv dM (3:2) 

With respect to the system S this flow work is negative at any point of 
entry, positive at points of exit from the system. 

Example 3:3. An open system simultaneously receives 1 lb of a substance having an 
equation of state as in Example 1:10 and, at a second point on its boundary, rejects 
the same weight of the same substance. At the point of entry the pressure and tem¬ 
perature of the system are, respectively, 50 psia and 300°F and remain constant at 
these values during the entry. At the point of exit the pressure and temperature are 
20 psia and 130°F and are also constant with time. Calculate (a) the flow work at 
entrance, (6) the flow work at exit, and (c) the net flow work per pound of fluid passing 
through the system. 

Solution: 

(a) At entry, P x = (20)(144) = 7200 psfa; v x = CT x /p x = (0.37)(760)/50 = 5.62 
ft 3 /lb. 

Flow work at entry = ~ J Q PiVidM = — (7200) (5.62) [ dM = —40,500 ft-lb/lb 

The negative sign is introduced to accord with the convention as to the sign of work, 

(b) At exit, P 2 = (20)(144) = 2880 psfa; y 2 = (0.37)(590)/20 = 10.9 ft 3 /lb. 

Flow work at exit = P 2 v 2 dM = (2880) (10.9)(1) = 31,400 ft-lb/lb 

(c) The net flow work = —40,500 + 31,400 = —9100 ft-lb per pound of fluid. 

3:4. The First Law and the Open System. As the system S of Fig. 3:1 

receives the element of fluid dM, its mass will change, as will also, in 
general, its total store of energy. This change in stored energy, according 
to the First Law as it applies to a process, must have derived directly from 
energy that crossed the system boundaries during the time required to 
introduce dM. This energy may be summarized to include work W that 
crossed the boundaries at points other than the point of entry of dM, flow 
work which accompanied the entry of dM, heat flow across any section of 
the boundary as dM was introduced, and the stored energy of dM itself 
as it was pushed across the boundary to become a part of system S. 
Expressing the change of stored energy of the system as being due to these 
effects in the form of an equation, we have 


THE FIRST LAW AND THE OPEN SYSTEM 


47 


dE = - ~ + PvdM + dQ+ (u + ^+jjdM 

- - - 7 - + dQ + + T + Wg + j) dM 

dW ( V 2 z\ 

= -- r + d Q + (h+- Tg + j)dM (3:3) 


3:5. Steady Flow. Many engineering applications of the open system 
will occur to the reader. For instance, in the case of the steam turbine the 
boundaries are fixed in position and are formed by the turbine casing and 
sections across the inlet and exhaust passages. Work crosses those 
boundaries along the turbine shaft (as shaft work), and heat may pass 
across them as due to radiation to the surroundings. Steam enters at the 
inlet and is discharged at lower pressure at the exhaust end of the turbine. 
This steam carries with it certain stored energy amounts and performs 
flow work on the system at entry, receiving flow work from the system as 
it is discharged. 

The use of Eq. (3:3) in analyzing such processes is greatly simplified if 
steady-flow conditions are assumed. This requires that: 

1 . The time rate of fluid entry (mass per unit time) is the same as that 
of exit from the system. 

2 . The fluid at entry has a uniform composition, state, and velocity 
that does not change with time. 

3. The same is also the case at exit. 

4 . The state of the fluid at any given point within the system does not 
change with time although different points within the boundaries may 
differ in their states. 

5 . The time rate of heat flow across the boundaries of the system is 
constant. 

6 . The same is true of the time rate of work flow across those boundaries. 

For the purpose of applying Eq. (3:3) to the steady-flow process, it is 

convenient to select a time interval during which 1 lb of the fluid will 
enter (and 1 lb leave) the system. Requirement 4 above assures that 
there will be no change in the energy stored in the system with time and 
therefore E 2 — Ei = 0. Designating the conditions at the entrance by 
the subscript i and those at exit by 2 , Eq. (3:3) becomes 


0 = - 


iW* , n , , , Vt 

+ 1 Q 2 + hi + 




J 


, zi _ h 

2Jg + J 2 2Jg 


Z2 

J 


(3:4) 


where iW 2 = work (other than flow work) which system performs on 
external systems during time required for unit weight of 
fluid to enter system, ft-lb 











48 


BASIC ENGINEERING THERMODYNAMICS 


iQ 2 = heat that system receives in same time interval, Btu 
h h h 2 = specific enthalpies of the fluid at entrance and at exit 
states, respectively 

t ~t 2 XT’ 2 

—E, J.A = specific stored energies due to motion of fluid entering and 
^ ^ leaving system, respectively 

z i, z 2 = specific stored energies due to elevation (gravity) at same 
points 

When Eq. (3:4) is rearranged, it assumes the typical form which is 
called the steady-flow energy equation, 


hi + 


zi 

2Jg ^ J 


Cl , ^ ^ , V 2 2 , Z 2 , 1 W 2 

+ - + iQ, - ht + w + 7 + -j- 


(3:5) 


The elevation terms, Zi/J and z 2 /J, are often dropped from this equation 
as representing energy amounts too small to be significant in comparison 
with the energy quantities represented by the other terms. In fact, 

differences in elevation between en¬ 
trance and exit are usually not stated 
even when it is intended that Eq. 
(3:5) is to be applied to the solution 
of the problem. Figure 3:2 is il¬ 
lustrative of the open-system steady- 
flow process. In accordance with 
the notation of Eq. (3:5), the entry 
section is designated as 1, the exit as 
2 . 



Example 3:5. Assume that steady-flow 
conditions existed between entrance and 
exit of the system described in Example 3:3, that the velocity at entrance was 100 fps, 
and that the point of entry was 5 ft above the exit. 


(а) Calculate the total energy introduced into the system at the point of entry per 
pound of fluid entering and the specific enthalpy of the entering fluid. 

(б) If the process was adiabatic and no external work was done, what was the velocity 
of the fluid at exit? 

( c) If the process was adiabatic and the velocity at exit was 500 fps, calculate the 
energy that was transferred to other systems as work per pound of fluid passing 
through the system, i.e., the external work per pound of fluid flow. 

(d) If no external work was done and the velocity at exit was the same as at entrance, 
how much heat was transferred per pound of fluid flow and in what direction? 

Solution: 

(a) ui = 16 + (0.46) (50) (5.62) = 145 Btu /lb. 

17,2 2l 1002 K 

61 = Ul + %Tg + J = 145 + (2) (778) (32.2) + 778 = 145 + 02 + °- 006 


= 145.2 Btu 




























THE FIRST LAW AND THE OPEN SYSTEM 


49 


It will be noted that this quantity is relative to the more or less arbitrary bases with 
reference to which u, V, and z are evaluated. For instance, in this case z has been 
referred to a datum level corresponding to the elevation of the exit section, and it 
will be noted that the stored energy due to elevation is so small that it has been 
dropped from the total as not significant. Naturally, if the stored energy at any 
other point in the system, such as the exit, is the subject of calculation, the same bases 
must be retained for the computation of each term. With respect to these datum levels, 
the total energy introduced into the system at the point of entry of the fluid is the 
sum of the energy stored in the pound of fluid that is entering and the (flow) work 
performed on the system in forcing its entry, or 145.2 -f (40,500/778) = 197.2 Btu 
per pound of fluid entering. The specific enthalpy of the fluid at entry is 

hi = 16 + 0.645pv = 16 + (0.645)(50)(5.62) = 197 Btu/lb [Ex. 2:9(/) 

( h) Applying Eq. (3:5), \Q 2 = 0 (the flow was adiabatic), and X IF 2 = 0 (no external 
work was performed). The potential energy terms Z\/J and z 2 /J may be dropped 
from the equation since Z\/J was insignificant and z 2 /J is zero. From (a), hi = 197 
Btu and Fi 2 /2 Jg = 0.2 Btu. Then 


F 2 2 


= 2Jg ( 


hi + 


or 


h 2 = 16 + (0.645)(20)(10.9) = 156.7 Btu 

Fi 2 n , iW t \ 

zp, + 1Q2 - A 2 - -J-) 

= (2) (778) (32.2) (197 + 0.2 + 0 - 156.7 - 0) 

= (50,000) (40.5) = 2,025,000 
V 2 = 1423 fps 

(c) 1 Q 2 = 0; hi = 197 Btu; fi 2 /2 Jg = 0.2 Btu; h 2 = 156.7 Btu. 

F 2 2 _ 500 2 _ R . 

2 Jg 50,000 

iW 2 = J (*1 + + 1 Q 2 -h 2 - = 778(197 + 0.2 + 0 - 156.7 - 5) 

= (778) (35.5) = 27,600 ft-lb 

The positive sign indicates that work (other than flow work) would have been trans¬ 
ferred to other systems as a result of this steady-flow process. 


V 2 2 F 1 2 


W :1V, - 0; ^ = Wg 


= 0.2 Btu. 


.«• - *• + u + ,J T - h ' - U - 156 ' 7 + °' 2 + 0 - 197 - °- 2 - 

The negative sign indicates that this heat passed out of the system. 


-40.3 Btu 


3:6. The Bernoulli equation is the steady-flow equation [Eq. (3:5)] 
as it applies to flow through a “ conservative ” system when no external 
(shaft) work is performed. For any real fluid, one of the piopeities that 
is associated with a given state is its viscosity. This property expresses 
the magnitude of the forces that are set up because of friction between the 
molecules due to their motion relative to each other. Because of vis¬ 
cosity, some shearing stresses will exist in any real fluid, but these stresses 











50 


BASIC ENGINEERING THERMODYNAMICS 


are sometimes so small that they may be ignored; this is especially true 
with respect to gases. A system composed of a hypothetical substance 
which does not have viscosity and in which these shearing stresses are 
therefore not present is called a conservative system. A conservative sys¬ 


tem could apply force to an adjacent system only in a direction normal to 
its boundaries; as it slides over the walls of the passage or across other 
portions of the fluid, an element of the fluid would not be retarded by 

friction. As it expanded in volume, the 
work performed by this element of fluid 
would, in the absence of friction, be a 
maximum. 

Let us consider a conservative system 
which consists of a unit mass of fluid as 
it passes through an open system in steady 
flow. The channel through which this 
flow takes place is pictured in Fig. 3:3, and 
it is our purpose to study the change in 
state of this unit mass as it moves between 
section 1 and section 2 of the channel. 
To aid in our study, we shall consider the 
segment of the passage that lies between 
sections a and b, which are an infinitesimal distance apart. Applying Eq. 
( 3 : 5 ) to this section of the passage b}^ expressing it in differential form, we 
have, when no external work (work other than flow work) is performed, 



Fig. 3:3. Steady flow and the 
conservative system. 


J dh + d 



+ dz — J dQ = 0 


Or, since Jh = Ju -f- Pv, 


J du + P dv + v dP — J dQ + d 



-f dz = 0 



The unit mass which we are studying constitutes a closed system. It is 
moving and therefore may change in velocity and in elevation, but veloc¬ 
ity and elevation are purely relative quantities, and an observer who was 
stationed “on board” the element and moved with it would see no change 
in these properties; to him, it would seem merely that external systems 
were changing in velocity and in height. For him, the system would be a 
simple system and one which, because it was a conservative system, could 
carry out only maximum-work processes. He would therefore apply Eq. 
(2:5) to an analysis of its behavior; in addition, he would be justified in 
substituting P dv for dW in that equation. Or, rearranging, 


J du + P dv — J dQ = 0 













THE FIRST LAW AND THE OPEN SYSTEM 


51 


Removing these terms from ( 1 ), we have 

v dP d ^ + dz — 0 (3:6) 

This is Bernoulli’s equation. Although it applies to a conservative sys¬ 
tem, it is often used to approximate flow relations for real fluids. 

Another concept that is occasionally of importance in engineering is 
that of the perfect fluid. The perfect fluid has all of the characteristics 
required of the fluid of a conservative system and, in addition, is incom¬ 
pressible. When Eq. (3: 6 ) is applied to an incompressible fluid, integra¬ 
tion gives 

V 2 

Pv + — + z = const ( 2 ) 

or, dividing by v , 

oV 2 

P H—t y- + wz = const . (3:7) 


in which p is the mass density of the fluid, in slugs per cubic foot, and w is 
its weight density, in pounds per cubic foot. The constant is the unvary¬ 
ing total head (total “pressure”) at all points along the channel of flow of a 
perfect fluid when no shaft work is performed. 

Bernoulli’s equation in the special form of Eq. (3:7) is often applied by 
the engineer to approximate flow relations for liquids, as in hydraulics, 
since they are characteristically compressible in only very slight degree. 
At times, when the changes in pressure are relatively very small, it is 
applied to the study of gas flows in order to simplify the mathematical 
treatment. 


Example 3:6. Water flows vertically downward in a converging tube. At a point 
100 ft below the entrance its velocity is 40 fps, and its pressure is 50 psig. At 500 ft 
below the entrance the pressure is 180 psig. Assume that the density of the water is 
constant at 62.5 lb/ft 3 and that this is a conservative system. Calculate the velocity 
at the lower section (a) by applying the Bernoulli equation and (6) by using the steady- 
flow energy equation. 

Solution: 


v = J_ = 0.016; dP = 144(180 - 50) = 18,720 psf; dz = -400 ft 
62.5 


(a) vdP +d (y) +dz = (0.016)(18,720) + d (y) 


400 = 0 


[Eq. (3:6)] 


d 


©- 


Vs 


v 2 2 


40 2 


2g/ 2 g 2 g 2 g 64.4 

V 2 2 = 64.4(100 + 24.8) = 8050 or ? 2 = 89.7 fps 


= 400 - (0.016) (18,720) = 100 


(b) Since this is a conservative system, dQ — du — P dv/J = 0. But dv = 0, and 
therefore dQ = du. If, in substitution in Eq. (3:5), any value of i Q 2 is substituted, 









52 


BASIC ENGINEERING THERMODYNAMICS 


it will be balanced by an equal difference between u 2 and u\ (which form a part of 
h 2 and hi) so that no effect will result in the calculation of V 2 , the required quantity. 
For simplicity, we shall assume that there is no heat flow and that consequently 
U 2 = Ui. In that case, 

A, -A, = - °- 16(18 °~ 50)(144) = 0.385 Btu 

J J 7/o 

Applying Eq. (3:5), 

VV = 2 Jg (h t - hi + + lQ> - -j-) 

/ 40 2 400 \ 

- 50 - 000 ( -°' 385 + 5W0 + 778 + ° " °) 

= 50,000(-0.385 + 0.032 + 0.514) = (50,000)(0.161) = 8050 
or 

V 2 = 89.7 fps 

An interesting feature of the solution of this example is the calculation of the change 
of enthalpy as A Pv/J. Expressed in the form of a differential this becomes dh = 
vdP/J. Examining further, it is noted that dh = d(u + Pv/J) = du -\- P dv/J 
+ vdP/J. For the maximum-work adiabatic process, dQ = 0 = du +P dv/J and 
therefore dh = v dP/J. In the case of an incompressible fluid (and most liquids are 
compressible only in negligible degree), this equation furnishes a simple method of 
calculating enthalpy changes during the maximum-work adiabatic process. The 
equation is valid, of course, even for gas systems that change in volume during this 
type of process but is less readily applied. 

3:7. The Continuity Equation of Steady Flow. In steady flow, the 
volume of flow passing any given section of the channel in unit time may 
be calculated either as the product of the mass rate of flow past that sec¬ 
tion and the (average) specific volume of the fluid as it passes the section 
or as the product of the cross-sectional area of the channel and the 
(average) velocity with which the fluid passes the section. Equating, 
Mv = AV, and, rearranging, M = AV/v. Since the flow is steady, 
requirement (a) of Art. 3:5 applies and M has the same value for all sec¬ 
tions along the channel. Thus we may write 


M = 


AV 

v 


AiVi 

Vl 


a 2 v 2 

v 2 


= const 


(3:8) 


in which the subscripts refer to successive sections normal to the direction 
of flow and A refers to the areas of those sections. Equation (3:8) is 
called the continuity equation of steady flow. It is often useful in providing 
a link between the velocity, the cross-sectional area of the passage, and 
those properties of the system which are known and may be used to enter 
the equation of state and solve it to obtain the specific volume. 

It is sometimes possible to select sections across all parts of which the 
velocity is the same. When an average velocity only is obtainable, it is 
customary to use this average velocity in the steady-flow energy equation 












THE FIRST LAW AND THE OPEN SYSTEM 


53 


although this procedure introduces some inaccuracy since the velocity is 
squared in the kinetic-energy term. If the error is considered too large 
to be ignored, the section may be subdivided into elementary areas over 
each of which the velocity varies negligibly and the kinetic energies of the 
respective elementary flows may be integrated. 

Example 3:7. (a) In part b of Example 3:5 assume a flow rate of 1 lb/sec, and 

calculate the necessary areas of the inlet and exit passages. ( b ) At one point between 
the entrance and the exit, the pressure of the fluid is 26.5 psia, and its temperature is 
173 F. What is the velocity at this point, and what is the necessary passage area 
for a flow rate as in part a? 

Solution: 


(a) From Eq. (3:8), 


Ax = 


M\Vi 


(1)(5.62) 

100 


M 2 V 2 = (1)(10.9) 
V 2 1423 


0.0562 ft 2 
0.0077 ft 2 


(b) This point in the flow will be designated by the subscript 0. Then 


Vo = 


CT 0 (0.37) (173 4- 460) 


= 8.85 ft 3 /lb 


Vo 26.5 

ho = 16 + 0.645p 0 vo = 16 + (0.645) (26.5) (8.85) = 167.3 Btu/lb 


[Ex. 2:9] 


Vo 1 = 2Jg (hi + + 1 Q 2 ~h 2 - = 50,000(197 + 0.2 + 0 - 167.3 - 


0) 


= (50,000) (29.9) = 1,495,000 or V 0 = 1230 fps 


A 0 = 


M 0 V 0 (1) (8.85) 


Vo 


1230 


= 0.0072 ft 2 


It will be observed that, although the values of the pressure, specific volume, tem¬ 
perature, and velocity at point 0 were intermediate between the values of the cor¬ 
responding properties at the points of entry into and exit from the open system, the 
area of the passage that confines the system and through which the flow is taking 
place is less at that point than at either entrance or exit. The methods used in solving 
this problem have an application to the design of nozzles, which are devices for utiliz¬ 
ing the stored energy of the fluid to give it increased velocity, and to the design of 
diffusers, which have the opposite effect. The basis of selection of the properties at 
the intermediate point 0 will be discussed in a later chapter of this text. 


3:8. The Steady-flow Heat Engine. The elementary steam-power- 

plant circuit diagramed in Fig. 3:4 is an excellent example of the steady- 
flow heat engine. Its efficiency as a heat engine may be obtained by 
applying either Eq. (2:14) or Eq. (2:15). As the fluid passes around the 
circuit, it may receive or discharge heat, perform or accept work, in 
devices that are each individually designed for the processes which are to 
be carried out in them. It thus does not suffer from the serious handicaps 
that existed in the case of the heat engine described in Chap. 2. Taken 
as a whole, the fluid enclosed within this steady-flow heat engine con¬ 
stitutes a closed system, but it differs from the type of closed system dis- 











54 


BASIC ENGINEERING THERMODYNAMICS 


cussed in Chap. 2 since its properties are not uniform throughout the sys¬ 
tem and since velocity may be an important property at certain points in 
the system. 

Although as a whole this type of heat engine constitutes a closed sys¬ 
tem, it may also be considered as an open-system cycle in the sense that an 
element of the fluid in making the circuit eventually returns to its initial 
state. The circuit may conveniently be divided into a number of open 
systems (in this case, four) for the purpose of examining the individual 
steady-flow processes that make up the complete cycle. 



Fig. 3:4. A steady-flow heat engine. 


In Fig. 3:4 the boiler and condenser are examples of the heat exchanger , 
which has as its object the heating or cooling of a fluid. When a heat 
exchanger is isolated as an open system, the differences in kinetic and 
potential energy of the fluid at entrance and at exit will ordinarily be 
negligible and no external work will be performed by the fluid in passing 
through the system. Applying the steady-flow energy equation, it then 
develops that the heat flow per pound of fluid entering (and leaving) is 

1 Q 2 = h 2 — hi (3:9) 

« 

in which hi and h 2 are the specific enthalpies of the fluid at entry and exit, 
respectively. 

The steady-flow energy equation may similarly be used to analyze the 
action of the prime mover. If that unit is a turbine, the boundaries of the 
open system, between sections across the inlet and exhaust passages, are 
fixed. Neglecting radiation (heat flow) and treating changes in kinetic 


















THE FIRST LAW AND THE OPEN SYSTEM 


55 


and potential stored energy between entrance and exit as negligible, it 
develops that 

iW 2 = J{h x - h 2 ) (3:10) 

in which iTT 2 is the external work (shaft work) per pound of fluid flow and 
hi and h 2 are the specific enthalpies of the fluid at entrance to and exit 
from the turbine, respectively. This equation, based on the First Law, 
does not limit the amount of work that may be done to any maximum 
amount for it makes no attempt to specify what relation shall exist 
between h 2 and hi in Eq. (3:10). Considering the relative preciousness 
of work and of heat, it will occur to the reader that some such limitation 
should apply. The discussion of this limitation will be left to a later 
chapter when we shall be better equipped to deal with the subject. 

The form of Eqs. (3:9) and (3:10) may cause the reader to gain the 
impression that enthalpy represents stored energy of the system. He is 
again reminded that this is not the case. Enthalpy is merely a composite 
property, a property that is of great usefulness as is indicated in these 
equations and others previously developed but not one that measures the 
storage of energy in the system. 

Any of the open systems that are discussed above may be subdivided 
into smaller systems for the purpose of more detailed analysis as seems 
desirable. For instance, in the case of the turbine a system may be 
selected that is appropriate for the study of the flow through a nozzle 
which is bounded by the walls of the nozzle and sections across the flow 
passages just ahead of and immediately after the nozzle. The change of 
kinetic energy will obviously be significant in this case, though the flow 
may be considered to be adiabatic and differences in elevation may be 
ignored. No external work is involved and the steady-flow energy equa¬ 
tion assumes the form 

Fo 2 Fi 2 

£ --g- = - AO (3:11) 

The left side of this equation is the change of kinetic energy, and since this 
is a form of energy that can be completely converted into work, it will 
again be evident that h 2 must be related to hi in some manner as yet 
undisclosed. 

3:9. Quasi-steady Flow. In the preceding article the steady-flow 
energy equation was applied to the turbine prime mover because it was 
the simplest type of prime mover for the purpose of demonstrating the 
steady-flow principle. If the prime mover is of the reciprocating-engine 
type, the boundaries of the open system that is formed are no longer fixed 
but pulsate periodically between certain limiting positions as shown in 
Fig. 3:5. Requirement 4 of Art. 3:5 is not satisfied though the state at 
any point in the open system does follow a definite change pattern and 




56 


BASIC ENGINEERING THERMODYNAMICS 


returns periodically to a given state. This is called quo,si-steady flow. 
The steady-flow energy equation may be applied to flow of this kind if 
receivers are provided as shown in Fig. 3:5 to damp out the oscillations in 
the state of the fluid at the entry and exit sections. Requirement 4 of 
Art. 3:5 is therefore not rigorous in limiting the application of the steady- 
flow energy equation. 



The external work is again shaft work as for the turbine but is in this 
case exchanged with other systems owing to a translation rather than a 
rotation of the shaft. The work delivery of the turbine was smooth, 
steady, and continuous, while iW 2 for the engine must be interpreted as 
the difference between the rate at which work leaves the system as its 
boundaries expand and the rate at which it reenters the system owing to 
the contraction that accompanies every pulsation. None the less, if the 
same assumptions are made (adiabatic flow, no significant differences in 
kinetic or potential energy between entrance and exit), it will be found 
that Eq. (3:10) will apply to the reciprocating engine as well as to the 
turbine prime mover. 

When the same methods are applied to the pump shown in the flow 
diagram of Fig. 3:4, the same assumptions will be found to apply and 
Eq. (3:10) will again develop. In this case, \W 2 being negative, it is 
evident that h 2 is greater than hi. 

Problems 

1. At entry to a receiver a gas has a velocity of 1000 fps, and at exit its velocity is 
100 fps. The exit is 10 ft above the entrance, and Ae is zero between entrance and 








































THE FIRST LAW AND THE OPEN SYSTEM 


r *7 

5 / 

exit states. What is the change of temperature between entrance and exit if the gas 
is (a) gas W ; ( b ) gas X ; (c) gas 7; (d) gas Z? 

2. A teakettle contains 5 lb of water and 0.01 lb of steam, both at 14.7 psia and 
212°F. The specific volume of the water is 0.016 ft 3 /lb, and its internal energy is 
180 Btu/lb. The similar values for the steam are 26.8 ft 3 /lb and 1077.6 Btu/lb. 
The kettle is placed on a stove, and heat is supplied until 1 lb of the water has boiled 
away, escaping with negligible velocity through the kettle spout to the atmosphere. 
What is the amount of external work? What is the amount of flow work? Neglect¬ 
ing the volume of the water which has boiled away, what is A E for the open system 
consisting of the kettle contents? If heat losses to the atmosphere through the kettle 
walls are neglected, what heat flow takes place? 

3. At entrance to a horizontal constricted passage, the pressure of a gas which is 
in steady flow through the channel is 100 psia, its temperature is 100°F, and its 
velocity is 100 fps. At the narrowest section, the pressure is 80 psia, and the tem¬ 
perature is 40°F. There is no heat flow through the walls of the channel between the 
two sections, nor is there any external work. What is the net flow work per pound of 
gas if it is (a) gas W ; (5) gas X ; (c) gas Y ; (d) gas Z? 

4. In Prob. 3, what is the velocity at the narrowest section of the passage, for 
each of the four gases? In each case, what is the ratio of the cross-sectional area of 
the passage at its narrowest section to its area at the entrance? 

5. At entrance to an open system, the pressure is 60 psia, the temperature is 
120°F, and the velocity is 800 fps. At exit, p 2 = 100 psia, t 2 = 260°F. The flow is 
steady, and the exit is 3 ft above the entrance. The gas is (a) gas W; (6) gas X; 
(c) gas Y; (d) gas Z. Answer the following questions: 

(1) What amount of stored energy is introduced into the system at the entrance per 
pound of gas crossing that boundary? 

(2) If the process is adiabatic and the velocity at exit is 200 fps, what is the external 
work per pound of gas flowing through the system? 

6. In steady slow (at negligible velocity) flow through a horizontal pipe, a fluid 
passes through an orifice plate containing a small opening. There is no external work 
and the flow is adiabatic. The entrance section is well upstream, and the exit section 
is well downstream from the orifice. The pressure and temperature at entrance are 
50 psia and 100°F; at exit the pressure is 20 psia. What is the temperature at exit 
if the fluid is (a) gas W; (b ) gas X; (c) gas F; ( d ) gas Z ? 

7. A gas is in steady flow through an open system. The temperature of the gas is 
600°F, and its velocity is 100 fps at entrance. The temperature at exit is 120°F, 
and the velocity is 300 fps. The flow is at the rate of 50 lb/min, and 10,000 Btu of 
heat leaves the system every hour. What horsepower does the system deliver as shaft 
work if the gas is (a) gas W; (b ) gas X; (c) gas Y; (d) gas Z? 

8. The temperature of an ideal gas is the same at entrance to and at exit from a 
passage, but the pressure drops from 75 to 25 psia. If the flow is steady and the 
velocities are the same at entrance and exit, what is the ratio of exit to entrance area? 

9. At the throat of a nozzle the velocity of a gas is 1500 fps, and its specific volume 
is 4 ft 3 /lb. The cross-sectional area at the throat is 1 in. 2 , and at the exit it is 2.5 
in. 2 The specific volume of the gas at exit is 16 ft 3 /lb. What is the velocity at exit? 

10. Liquid water (density = 62.4 lb /ft 3 ) flows adiabatically through a horizontal 
converging tube. At a certain section its pressure is 100 psia, and its velocity is 20 
fps. At a later section the velocity has become 50 fps. Assuming this to be a con¬ 
servative system, what is the pressure at the second section? 

11. In Prob. 10, if the second section had been 100 ft below the first, what would 
have been the pressure at that section? 

12. Compressed air flows in an open line, horizontal and of constant diameter. At a 


58 


BASIC ENGINEERING THERMODYNAMICS 


section near the entrance, the pressure of the air is 90 psia, its specific volume is 
2.5 ft 3 /lb, and its velocity is 100 fps. At a later section, the pressure is 85 psia, and 
the specific volume is 2.3 ft 3 /lb. The specific internal energy decreased by 13.3 Btu 
between the sections. What heat flow (Btu per pound of air carried) took place 
between the two sections, and in what direction did it flow? 

13. A boiler evaporates 50,000 lb of water per hour. At a section of the feedwater 
line entering the boiler, the enthalpy of the water is 68 Btu/lb, the velocity is 30 fps, 
and the elevation is 5 ft above the boiler-room floor. At a section of the steam line 
leaving the boiler, the enthalpy of the steam is 1200 Btu/lb, the velocity is 150 fps, 
and the elevation is 30 ft. If 75 per cent of the heat value of the coal (12,000 Btu per 
pound of coal) is used in heating the water and steam, how many pounds of coal are 
burned per hour? 

14. A steam turbine receives steam having a specific enthalpy of 1250 Btu/lb. 
The rate of steam flow is 12,000 lb /hr, and the turbine delivers 800 kw. The heat loss 
by radiation may be considered negligible. The steam enters the turbine at 100 fps 
and leaves at a velocity of 500 fps. The exit section is 5 ft below the entrance. What 
is the enthalpy of the steam at exit from the turbine? 

15. A steam engine receives 3000 lb of steam per hour at an enthalpy of 1195 
Btu/lb and exhausts steam having an enthalpy of 1080 Btu/lb. The heat loss from 
the engine cylinder and the piping between the entry and exit sections is at the rate of 
6000 Btu/hr. Assuming that the velocities and elevations at entrance and exit differ 
negligibly, what power does this open system develop? 

16. The enthalpy of the steam as it enters a condenser is 1080 Btu/lb; that of the 
condensate at exit is 150 Btu/lb. Three thousand pounds of steam is condensed per 
hour. The velocity of the steam at entrance is 300 fps; of the condensate it is 30 fps. 
If the cooling water increases in temperature by 50°F in flowing through the condenser 
and 50,000 Btu of heat is radiated per hour directly to the atmosphere from the shell 
of the condenser, how many pounds of cooling water must be supplied per hour? 
Neglect differences in elevation. The specific heat of water is 1. 

17. Derive an expression for the work required to operate a feedwater pump, per 
pound of water pumped, by applying Eq. (3:5). Assume inlet and outlet velocities 
and elevations equal, no heat transfer, that the system is a conservative system, and 
that the specific volume of the water remains constant as the pressure increases. 

18. Steam flows through a small opening from a region of higher pressure to a cham¬ 
ber in which a lower pressure is maintained. If sections are chosen at a suitable 
distance on the two sides of the opening and the flow rate is small, the difference 
between upstream and downstream kinetic energy is small and may be ignored. If 
the flow is adiabatic, show that no change of enthalpy will result. 

19. The specific enthalpy of the air as it leaves an air compressor is 47 Btu greater 
than as it entered. One thousand pounds of air is compressed per hour, 27 hp being 
needed to drive the compressor. Neglecting differences of velocity and elevation 
between entrance and exit, calculate the heat rejected to the cooling water jacket and 
the atmosphere per pound of air compressed. 

20. One pound of H 2 0 flows around the circuit which is diagramed in Fig. 3:4. 
In the boiler it receives 850 Btu, in the condenser it gives up 700 Btu of heat. Radia¬ 
tion from the prime mover, the feedwater pump, and the various steam and water 
lines accounts for a heat loss of 25 Btu to the atmosphere. Work in the amount of 
1 Btu is supplied to drive the feedwater pump as it pumps the pound of condensate 
into the boiler. What is the work output by the prime mover per pound of steam 
flow? What is the net work output of the heat engine per pound of flow? What is 
the efficiency of this steady-flow heat engine? 


THE FIRST LAW AND THE OPEN SYSTEM 


59 


Symbols 

a acceleration 

A cross-sectional area of a channel, normal to the direction of flow 
e stored energy of unit mass 
E stored energy of a system 
F force 

g acceleration due to gravity 
h specific enthalpy 
J proportionality factor 
l distance 
m mass, slugs 
M mass rate of flow 
p pressure, psi 
P pressure; pressure, psf 
Q heat flow per unit-mass rate of flow 
l time 

u specific internal energy 
v specific volume 
V velocity 
w weight density 

W shaft work per unit-mass rate of flow 
z height; elevation above a datum level 

Greek Letters 
p mass density 


CHAPTER 4 


THE REVERSIBLE PROCESS AND THE REVERSIBLE CYCLE 

4:1. Thermodynamic Reversibility. Continued reference has been 
made in these pages to the fact that work is a more valuable form of 
energy than is heat. The reader can scarcely have failed to arrive at the 
same conclusion, based on his own independent observation. Any 
process during which work is used to produce an effect that could have 
been produced, at least in part, by a flow of heat therefore falls short of 
thermodynamic perfection and represents a waste of energy in its more 
valuable form. 

If, after the conclusion of a process, the system may be restored to its 
original state and all other systems returned to their original conditions 
as at the beginning of the original process, it may reasonably be concluded 
that no such wastage accompanied that process. In this case the process 
is said to be thermodynamically reversible. Reversibility therefore pro¬ 
vides a test that can be applied to determine whether or not the process 
has been perfectly executed in the thermodynamic sense and thus to 
prove whether work or any form of energy which could have been 
changed completely into work (such as kinetic or potential mechanical 
energy, for example) has been used for purposes that could have been 
served by a flow of heat. 

4:2. Reversibility and the Closed-system Process. Returning to the 

statement of what constitutes reversibility in a process, it is noted that 
not only must the system be restored to its original state but all other 
systems that were involved in the process must be returned to their 
original condition. This requires that: 

1. If an external system receives work during a reversible process, it must 
store the amount of energy equivalent to this work in some form such that it is 
available for return as workflow in the opposite direction and in undiminished 
amount during the contemplated return of all systems to their original states. 
As an example of how this might be done, consider that the work per¬ 
formed on a piston by the expanding system could be used to give angular 
acceleration to a flywheel and thus to store kinetic mechanical energy in 
the external system, which here consists of the piston, flywheel, and con¬ 
necting linkage. During the return process, work could pass from this 
external system as the flywheel decelerates. If all parts of this external 
system were frictionless, the work flow in the two directions would be the 
same and this requirement would be satisfied. 

60 


THE REVERSIBLE PROCESS AND REVERSIBLE CYCLE 61 


2. One of the requirements for the reversibility of a process is that it 
may be stopped at any point and the return to the original state for all 
systems involved made from that point. Therefore, for any reversible 
process the path followed during the return process must retrace that traversed 
in the original process. 

3. It must require exactly as much work from external systems to restore the 
system to its original state as was delivered to those external systems during 
the original process. This means that the process must be a maximum- 
work process if the system is to be returned to its original state over the 
same path. Otherwise (a) less work than is represented by the area 
under the path on a PV diagram would be delivered by the system to 
external systems as the system expanded, ( b ) this area would represent the 
minimum amount of work (in the absolute sense) which these external 
systems would be called upon to return to the system in retracing the 
same path, and (c) an inequality would result. Maximum work requires 
that there shall be no degree of unrestrained expansion and no paddle- 
wheel work (friction) during a closed-system process. 

4. When heat flows to an external system during the process, that external 
system must store the equivalent amount of energy in a form and under condi¬ 
tions such that it can be returned as heat and in its full amount during the 
return process. Since heat flows only in response to a differential of tem¬ 
perature and always from the system at higher to that at lower tempera¬ 
ture, it is clear that no process involving the flow of heat can be truly 
reversible. However, it may approach reversibility as a limit as the 
differential of temperature between the systems which exchange heat 
decreases. 

A review of the preceding paragraphs will indicate that friction, unre¬ 
strained expansion, and the flow of heat across a finite temperature interval 
will prevent the attainment of reversibility in a closed-system process. 
These factors are always present in any real process, and therefore no real 
process can be completely reversible and thus perfectly executed in the 
thermodynamic sense. But as the effects of friction are reduced toward 
zero by improved methods of lubrication, as the effects of unrestrained 
expansion are minimized by slow piston movement, and as the introduc¬ 
tion of better conductors makes it possible to transfer heat at the required 
rate with vanishing temperature differential, the ideal process develops 
as a limit which it is possible to approach. The reversible process is 
valuable in establishing a standard that cannot, even in theory, be 
exceeded. 

4:3. The Reversible Adiabatic Closed-system Process. Since no 
heat flow is involved, reversibility can be conceived in an adiabatic proc¬ 
ess with less exercise of the imagination than is necessary for the other 
processes described in Chap. 2. It is also the type of change for which 


62 


BASIC ENGINEERING THERMODYNAMICS 


the real process can be made to approach the ideal reversible process most 
closely for the same reason. The path followed by the reversible adia¬ 
batic is necessarily a maximum-work path (see Art. 2:12) over which 
dU/dV = —P/J. On pressure-volume coordinates this path would 
have, for any gas, somewhat the appearance of the path 1-2 in Fig. 4:1, 
which also shows the cylinder in which the expansion is pictured as taking 
place and the piston that forms a part of the external system which 
receives work during the expansion. The flywheel and connecting linkage 

are not shown. The walls and head of 
the cylinder are insulated to prevent 
the passage of heat. The piston, fly¬ 
wheel, and linkage are assumed fric¬ 
tionless, and the inertia of the flywheel 
is large so that only slow piston move¬ 
ment is attained in response to the 
force applied by the expanding system. 

Our first assumption, for simplifica¬ 
tion, will be that the cylinder is located 
in a void so that all of the work de¬ 
livered by the expanding gas system 
will be utilized in increasing the kinetic 
energy of the piston, flywheel, and 
linkage. Because of the slow piston 
movement that has been specified, the 
resistance offered by this external sys¬ 
tem is only infinitesimally less than the 
product of piston area and the pressure 
of the gas system against its stationary 
boundaries (the pressure plotted along 
the path 1-2). As the piston moves from one dead-center position to the 
other, the work delivered to it is accordingly only infinitesimally less than 
the area under the path 1-2 on the PV diagram. 

As the piston reaches the outer end of its stroke (point 2), the flywheel 
has its greatest store of kinetic energy. As the dead-center position is 
passed, the piston is moved back to the left at the expense of the stored 
energy of the flywheel until the force applied by the piston becomes 
infinitesimally greater than the resistance offered by the pressure of the 
gas system. Compression will then proceed over the system path 2-1 
until the original state has been reached. The work required of the 
piston-flywheel external system during this return process is seen to be 
infinitesimally greater than that which it received during the expansion of 
the gas system, but this difference vanishes as the rate of piston movement 
decreases. Thus, as a limit, the flywheel will return to its original state 



r 

Wmm. 

-Po 1 

i 

i 

i 

i 

i 

i 

i 

i 

i 

i 

w////mA 

_ i 

i 

i 

i 

imamm 



Fig. 4:1. The reversible adiabatic 
process. 



















THE REVERSIBLE PROCESS AND REVERSIBLE CYCLE 63 


(which would include velocity as one of the properties) at the same 
time the gas system again reaches state 1. The process, as conceived 
and under the conditions that have been assumed, is declared to be 
reversible. 

Now consider that the operation as described above did not take place 
in a void but in the midst of a vast medium having the pressure P o and the 
temperature To. Since the gas system is thermally insulated, the tem¬ 
perature of the surroundings will have no effect upon its expansion or 
compression path. However, during expansion of the gas system a part 
of the work performed by this system must be expended in pushing the 
medium back against the resistance offered by its pressure as it acts 
against the rear face of the piston; this portion of the total work is repre¬ 
sented by the area below the line P 0 on the diagram and between the vol¬ 
ume limits of the expansion. The balance of the work delivered by the 
gas system as it expands (the area above P 0 and under the path 1-2) is all 
that is available for accelerating the piston-flywheel system, and the 
motion of the piston is thus slower than before. During the return stroke, 
this smaller storage of energy in the flywheel will none the less be sufficient 
to return the gas system to state 
1, for on this stroke the medium 
will return work to that system 
equal to the amount received dur¬ 
ing the expansion. The process 
still conforms to our concept of 
reversibility, the only essential 
difference being that two external 
systems are involved instead of 
one. 

4:4. The Reversible Isother¬ 
mal Process. The simplest case 
that involves heat flow between 
the system and its surroundings 
develops when those surroundings 
remain at constant temperature 
during the process. This case is 
illustrated as it applies to a re¬ 
versible process in Fig. 4:2. The 
walls of the cylinder are noncon¬ 
ducting as in Fig. 4:1, but the head is a perfect conductor of heat, and an 
external system S' that is capable of supplying or receiving heat without 
change of its temperature T s is in contact with this head. Thus the 
temperature of the system enclosed within the cylinder and behind the 
piston may vary only infinitesimally from that of S'. A second external 



Fig. 4:2. The reversible isothermal proc¬ 
ess. 






















64 


BASIC ENGINEERING THERMODYNAMICS 


system consisting of a frictionless piston-flywheel mechanism is provided 
to receive work and return it as before. 

At the beginning of the process, the enclosed system has a state corre¬ 
sponding to state 1 on the pressure-volume diagram. The temperature 
Ti is identical with that of the external system S'. As the enclosed sys¬ 
tem undergoes the first infinitesimal stage of its expansion, there will be no 
flow of heat from S' until the temperature of the system has dropped 
infinitesimally below Ts. After that first step in the expansion, however, 
a flow of heat from S' will begin that will be sufficient to maintain the tem¬ 
perature of the enclosed system at Ts — dT and the path 1-2 will be an 
isothermal at this temperature level. As the process proceeds to its end 
point 2 (the end of the piston stroke), work is done on the piston-flywheel 
external system and the kinetic energy of that system is increased. 

The direction of piston movement is now reversed, and the flywheel 
gives back some of its store of kinetic energy as the enclosed system is 
compressed. During the first infinitesimal stage of that compression the 
temperature of the enclosed system is raised to Ts + dT, after which it is 
maintained at this level as the system returns to state 1 over a path that 
is only infinitesimally above the path traced during the outward move¬ 
ment of the piston. As it retraces its path, the enclosed system may now, 
because of the favorable temperature differential which has been created, 
return to S' the same amount of heat that it received during the expansion. 
Neglecting differences of infinitesimal amount, both external systems have 
been returned to their original condition as the enclosed system reaches 
its original state, and the process may be classified as thermodynamically 
reversible. 

4:5. Reversible Heat Flow at Variable Temperature. When the 

demonstration of a process requires both a flow of work across the bounda¬ 
ries of a system and the flow of heat across those boundaries while the tem¬ 
perature of the system changes finitely, the application of the concept of 
reversibility to that process becomes still more involved. The constant- 
pressure process illustrated in Fig. 4:3 is an example, though any variable- 
temperature path during which a change in system volume took place 
would also serve. 

To conceive of this process as reversible requires the same cylinder with 
nonconducting walls and a head that is a perfect conductor of heat which 
was described in the preceding article. Instead of a single external sys¬ 
tem with which heat is exchanged, however, a series of such systems are 
placed in contact with the head of the cylinder, and the expansion of the 
enclosed system proceeds as a series of minute isothermals, the tempera¬ 
ture at which each is carried out differing only infinitesimally from that of 
adjacent path segments. In retracing the process the external systems 
S' to S n are brought into contact with the cylinder head in reverse order. 


THE REVERSIBLE PROCESS AND REVERSIBLE CYCLE 65 


As the temperature differential between successive systems in the series 
$ to S n decreases toward zero (and the number of individual systems in 
the series consequently increases toward infinity), the process will 
approach reversibility as a limit. 

4.6. The Constant-volume Reversible Process. If the external sys¬ 
tem that supplies the heat necessary to demonstrate the constant-volume 
process, as described in Chap. 2, is at constant temperature, then that 
process is irreversible since the temperature of the enclosed system is not 


P 

Pt>p2 


Po 


V, V 2 V 

Fig. 4-3. The reversible constant-pressure process. 

such that it can return this heat. If, however, a series of external systems 
is provided as suggested in the preceding article, the process may approach 
reversibility as a limit. 

Consider the arrangement sketched in Fig. 4:4. The cylinder A is 
nonconducting and is fitted with a piston that is leakproof and is a part of 
a frictionless mechanism including a flywheel and the necessary linkage. 
A passage, the walls of which are also nonconducting, connects the oppo¬ 
site heads of the cylinder. This passage is divided into compartments 
separated by partitions that are nonconducting to heat but will permit the 
slow flow of fluid. Each compartment contains a capacitor that is 
capable of receiving and emitting heat; the reader may visualize these 
capacitors as consisting of metallic but porous material such as iron or 
copper filings. As we first examine the cylinder and passage, we find that 
the following observations apply: 


+ 



|! 

K 

+ 


+ 




I' 




C 


c 














/ 2 






























66 


BASIC ENGINEERING THERMODYNAMICS 


1. The piston is at the bottom of the cylinder, and the air above it is at 
a uniform temperature TV 

2. The capacitors vary in temperature from 7\ + dT to T 2 T* dT from 
the top to the bottom of the passage, and each capacitor differs in tem¬ 
perature from adjacent capacitors by dT. 

3. The thermal capacity of each capacitor is the same as that of the 
entire fluid system which is confined within the cylinder. That is, the 

amount of heat that, when removed from 
the capacitor, will cause its temperature 
to decrease by dT will be sufficient to in¬ 
crease the temperature of the fluid system 
by dT. 

4. The flywheel is assumed to be in slow 
motion. 

Under the influence of the flywheel the 
piston will move upward in the cylinder. 
As it does so, the flywheel is not slowed, 
however, for the motion of the piston is so 
slow that the pressure on its upper and 
lower faces may be considered to differ 
only infinitesimally. When the piston has 
completed its upward stroke, the fluid will 
have been transferred to a location below 
it. As a result of heat received from the 
capacitors the temperature of the fluid is 
now T 2 , and the temperature of each capac¬ 
itor is lower than its original temperature 
by dT. 

As the piston begins its downward movement, the direction of fluid flow 
through the capacitors is reversed and, when the piston is again at the 
bottom of the cylinder, the temperature of the fluid will, neglecting differ¬ 
ences of small order, be the same as it was originally; the same will be true 
of the temperatures of the capacitors. The motion of the flywheel has 
remained steady throughout, and the system of which it is a part has 
therefore returned to its original condition. No other systems have been 
affected. 

The foregoing is an example of one method by which the constant- 
volume process may be conceived as taking place reversibly in the limit. 
This method will later have special interest to us since it illustrates how 
the same amount of heat that the system gave up as its temperature was 
reduced from a higher to a lower temperature might be made available for 
eventual return to the system with the purpose of increasing its tempera¬ 
ture between the same limits of temperature. The reader will note that the 



Fig. 4:4. The reversible con 
stant-volume process. 

























THE REVERSIBLE PROCESS AND REVERSIBLE CYCLE 67 


return of this heat would not be limited to cases in which the same path 
was followed in reverse but that it would also be available for return to the 
system over any constant-volume path that took place between the same 
temperature limits. Thus a proposed cycle may include two constant- 
volume processes that take place at different system volumes but for one 
of which the system temperature decreases from T 2 to T\ while for the 
other the temperature increases from 7\ to T 2 . In that case the heat 
rejected during the first process may provide the heat supply during the 
second. In Eqs. (2:14) and (2:15) neither Q s nor Q R would include heat 
exchanged between the parts of the cycle in this manner, and reference to 
those equations will indicate that the efficiency of the cycle would be cor¬ 
respondingly increased. When processes are paired in this manner, the 
mutual exchange of heat is called regeneration. Regeneration may be 
applied between pairs of constant-pressure processes as well as those 
which take place at constant volume. 

4:7. External and Internal Reversibility. The three factors any of 
which will always account for the irreversibility of a process have been 
shown to be friction, a finite pressure differential, and a finite temperature 
differential. In setting up the conditions under which each of the proc¬ 
esses analyzed in the foregoing articles may approach reversibility, these 
factors are eliminated at the boundaries of the system. The processes may 
therefore be classified as externally reversible. If any of these factors 
operate within the boundaries of the system, the result is classified as 
internal irreversibility. If the system is not in equilibrium, a tendency 
will exist toward an internal change that will be irreversible in character. 

For example, suppose the system to consist of two parts which differ 
finitely in temperature. The reader will realize that if these two parts are 
allowed to mix or even to exchange heat without intermingling, the opera¬ 
tion is irreversible in the thermodynamic sense, for the system cannot be 
returned to its original state without calling on external systems for 
assistance and leaving a record in terms of the changed states of these 
external systems. The same will be true if the system is not in equilib¬ 
rium by reason of a finite pressure differential between its parts and if this 
pressure difference is allowed to adjust itself within the system. Relative 
movement between two or more parts of the system which is accompanied 
by friction will also produce an internal irreversibility. The further 
stipulation as a condition for complete reversibility, external and internal, 
must therefore be made that the path followed during any reversible process 
must connect equilibrium states of the system. 

Internal irreversibility may sometimes be eliminated by redefining the 
system in terms of new boundaries which will enclose only those parts of 
the original system which are in mutual equilibrium. All other parts 
thereafter are classified as external systems. 


68 


BASIC ENGINEERING THERMODYNAMICS 


4:8. Reversibility in Steady Flow. A steady-flow process is reversible 
if the stream of fluid can be restored to its original state (including velocity 
and elevation as properties which assist in defining that state) by any 
means which will leave no history of the process in the changed states of 
external systems. An example is the steady adiabatic flow of a fluid 
through a frictionless nozzle and diffuser as illustrated in Fig. 4:5. Dur¬ 
ing flow through the nozzle (from section 1 to section 2) the kinetic energy 
of the fluid increases by an amount equal to the decrease of enthalpy [see 
Eq. (3:11)]. In the diffuser (section 2 to section 3) exactly the opposite 
effect is caused, returning the stream to its original state without affecting 
any external system. This combination of nozzle and diffuser is called a 
venturi. In no real venturi will the increase of pressure between sections 

2 and 3 be quite equal to the de¬ 
crease that has taken place between 
sections 1 and 2. This reflects the 
„ effects of friction which enters into 
the picture in increased degree as 
the velocity mounts. The revers¬ 
ible process continues for steady 
flow to be a limiting process. 

Irreversibility in steady flow is occasioned by the same three factors 
that apply in the closed-system process. When none of these is present, 
the process can be presumed to be reversible. Thus adiabatic flow 
through a turbine may be assumed to be reversible if the flow is friction¬ 
less and if no unrestrained expansion of the fluid has taken place, for the 
same work which was delivered by the turbine will be sufficient to return 
the stream of fluid to its original state through the agency of a frictionless 
compressor. No external system will have been affected. The path 
followed in both the turbine and the compressor will be that described in 
Art. 2:12. 



Fig. 4:5. Reversible steady flow. 


Example 4:8. A gas having equations of state as in Example 1:10 enters a reversible 
adiabatic turbine at a pressure of 150 psig and a temperature of 1200°F. It is 
exhausted from the turbine at atmospheric pressure, (a) What is the exhaust tem¬ 
perature? (6) How much power does the turbine develop per pound of gas flow per 
second? 

Solution: 

(a) V x = 150 + 14.7 = 164.7 psia; Tx = 1200 + 460 = 1660°R; p 2 = 14.7 psia; 


k = 1.4 (see Example 2:125); vx = 


CTx (0.37) (1660) 


V i 


164.7 


= 3.73 ft 3 /lb. 


/ r),\Wk /164 7V/1.4 

ViVi k = P*V 2 k or v 2 = Vl = 3.73 f ) = (3.73)(5.63) = 21.0 ft 3 /lb 


T* = = (14.7) (21.0) = 833°R, or 373°F 
















THE REVERSIBLE PROCESS AND REVERSIBLE CYCLE 69 


(b) hr = 16 + 0.645piVi = 16 + (0.645)(164.7)(3.73) = 413 Btu/lb 
h 2 = 16 + (0.645)(14.7)(21.0) = 215 Btu/lb 
iW 2 = J{h x - h 2 ) = 778(413 - 215) = 154,000 ft-lb/lb [Eq. (3:10)] 

The horsepower per pound of gas flow per second is 154,000/550 = 280 hp. 


4:9. The Carnot Engine. By definition, the heat engine operates 
continuously. For continuous operation some sort of cycle of operations 



Fig. 4:6. The Carnot cycle (Example 
4:8.) 




X 

C3 




X 

o 

"K 

s 

-s 

$ 





Fig. 4:7. The Carnot engine. 


is required. It would seem probable that, for most effective performance 
of its function, the heat engine should operate on a cycle which consists 
exclusively of processes that are perfectly executed in the thermodynamic 
sense, i.e., reversible. A cycle which is entirely composed of processes 
that are reversible is a reversible cycle. 

In 1824 Sadi Carnot, a French scientist, proposed a reversible cycle 
during which heat would be exchanged only with a source at higher and a 
sink, or refrigerator, at lower temperature. His cycle, illustrated in Fig. 
4:6 for a gas that has an equation of state similar to Eq. (1:6), consists of 
an isothermal expansion at the temperature Ts of the source and an 
isothermal compression at the lower temperature T R of the refrigerator. 
These two isothermals are connected by two adiabatic processes, an 
adiabatic expansion which begins at the end of the isothermal expansion 
lowering the temperature of the system from T s to T R and an adiabatic 
compression that follows the isothermal compression and again raises the 
temperature of the system to T s . 

The Carnot engine operates on the cycle described above. A possible 
arrangement is diagramed in Fig. 4:7. A frictionless piston-flywheel 
system is provided as the agency for work delivery. The walls of the 
cylinder are nonconducting, but the head is a perfect conductor of heat. 
Provision is made for bringing instantaneously into contact with the head 
either the source at T s or the refrigerator at T R ; the head may also be 






















70 


BASIC ENGINEERING THERMODYNAMICS 


covered by a nonconducting cap. The first stage of the expansion is 
carried out with the source at Ts placed in contact with the head of the 
cylinder; the path 1-2, a reversible isothermal during which heat is 
received from the source at temperature Ts, results. When point 2 is 
reached, the source is instantaneously replaced by the nonconducting cap 
and the expansion continues to the end of the piston stroke; the path 2-3 
is a reversible adiabatic during which the temperature of the system 
decreases to that of the refrigerator. At the instant that the piston 
reverses the direction of its motion (point 3 in the cycle) the refrigerator 
at Tr replaces the nonconducting cap. The first stage of the compression 
(3-4) therefore develops as a reversible isothermal compression at Tr 
during which heat is rejected to the refrigerator. The cycle is closed over 
path 4-1, which requires that, as the piston reaches point 4 of the cycle, 
the refrigerator is replaced by the nonconducting cap at the head of the 
cylinder and the rest of the stroke of the piston is carried out as a reversi¬ 
ble adiabatic compression of the enclosed system. The only step then 
necessary to prepare for a retraversal of the cycle is to remove the non¬ 
conducting cap and move the source into position at the head of the 
cylinder. 

Reviewing the characteristics of this engine and the cycle on which it 
operates it is noted that: 

1. Heat is received only from a source at constant temperature and only 
during the process 1-2. To conform with the notation of Eqs. (2:14) and 
(2:15), the amount of heat transfer during this portion of the cycle may 
be designated as Q s . 

2. Heat is rejected only to a refrigerator at constant temperature and 
only during process 3-4. The amount of heat so transferred is Qr [see 
Eq. (2:15)]. 

3. Since all processes are maximum-work processes, a net dividend of 
work, as represented by the area enclosed within the cycle, results from 
the traversal of the cycle. This amount of work must, by the First Law, 
equal Q s — Qr, and Q s is therefore larger than Q R . 

4. Provision could be made for a lowered temperature T R of the refrig¬ 
erator with Ts unchanged by extending process 2-3 until this lower tem¬ 
perature was reached. This would also result in a lowering of the path 
3-4 and a consequent increase in the enclosed area of the cycle and thus in 
the difference Qs — Qr. But process 1-2 need not be affected, Q s would 
thus have the same value as before, and therefore Qr must have decreased 
as T r was reduced; the efficiency of the Carnot heat engine must have 
correspondingly increased [see Eqs. (2:14) and (2:15)]. 

5. Since all the processes which comprise the cycle on which its opera¬ 
tion is based are individually reversible, the Carnot engine may follow 
them in reverse order (counterclockwise in Fig. 4:6 instead of clockwise). 


THE REVERSIBLE PROCESS AND REVERSIBLE CYCLE 71 


In this case, as will be shown below, the Carnot engine will become a 
refrigerating machine, or heat pump. 

In analyzing the action of the Carnot engine as a heat pump the term 
hot body will replace source in order to avoid confusion since it will develop 
that the high-temperature reservoir no longer supplies heat to the system 
but instead receives it from the system. The term refrigerator is still 
appropriate since it carries no implication of the direction of heat flow but 
does indicate that the temperature of that reservoir is relatively low. 

In reversed traversal of the cycle we shall start, for convenience, at 
point 4 with the refrigerator at Tr in contact with the head of the cylinder. 
The process 4-3 is now an expansion during which the heat flow is the same 
in absolute amount, though opposite in direction, as for the process 3-4 in 
clockwise movement around the cycle. At point 3 the nonconducting 
cap replaces the refrigerator, and, during the reversible adiabatic com¬ 
pression 3-2, the temperature of the system rises to that of the hot body 
( Ts ). At point 2 the nonconducting cap is removed, the hot body placed 
in position, and an isothermal compression 2-1 at Ts thereafter proceeds. 
Evidently the heat transfer for this process is the same in amount ( Q s ) as 
for process 1-2 though now the heat flows from the system into the hot 
body. When point 1 is reached, the nonconducting cap is again placed 
in position and the cycle is closed by the reversible adiabatic process 1-4. 

Comparing the operation of the Carnot heat engine with that of the 
Carnot heat pump which operates on the same cycle but in the reverse 
direction, it is seen that neither the amount of heat exchange with each of 
these reservoirs nor the net work exchanged with other external systems 
has changed; the direction of energy flow alone differs. The term reversi¬ 
ble engine thus assumes a greater significance than simply that of an 
engine that operates on a cycle composed entirely of reversible processes. 
For a reversible engine, after operation for a period as an engine, could be 
reversed to travel its cycle in the opposite direction (as a heat pump) and 
would be capable of restoring to its source all of the heat that it had 
received from that source, at the expense only of heat taken from the 
refrigerator which had, during operation as an engine, been deposited 
with that refrigerator, and of work returned by other systems with which 
it had been deposited. Thus the engine is reversible in the sense that it 
can restore all systems to their original states and leave no history of its 
operation on them. 

There are no real Carnot engines. A review of the conditions that sur¬ 
round the operation of such devices will be sufficient indication to the 
reader as to why they are not practical devices. Their absence from the 
real power plant does not change the value of the Carnot-engine concept 
as a device which serves as the basis of interesting speculation in the field 
of thermodynamics. 


72 


BASIC ENGINEERING THERMODYNAMICS 


The Carnot engine has in this article been described and analyzed in 
its classic form, with respect to its operation on a closed-system cycle. 
With suitable changes in apparatus and the introduction of flow around a 
closed circuit, an equivalent open-system cycle could be devised; for the 
present, this is left to the reader to demonstrate. 


Example 4:9. The system contained within a Carnot engine (the working sub¬ 
stance) consists of 1 lb of the gas for which equations of state are as in Example 1:10. 
Referring to the notation of Fig. 4:6, V\ = 3 ft 3 , V 2 = 6 ft 3 , ts = 500°F, and t,R 
= 100°F. (a) Calculate the pressure, volume, and temperature at points 1, 2, 3, 

and 4 around the cycle. (6) Calculate Qs and Qr per cycle and, by difference, the 
net work of the cycle, (c) Calculate the work which accompanies each of the proc¬ 
esses, and obtain the net work of the cycle as the total. ( d ) Compare the net work 
of the cycle with the gross work of expansion (the sum of the work performed during 
processes 1-2 and 2-3). ( e ) What is the efficiency of the cycle? 

Solution: 

(a) Point 1: v x = Vi = 3 ft 3 ; T x = T s = 960°R; Pl = CTl - (°- 3 ')( 96 °) 


Vi 


Point 2: v 2 = V 2 = 6 ft 3 ; T 2 = T s = 960°R; p 2 = 


CT 2 (0.37) (960) 


v 2 


6 


Point 3: T 3 = T R = 560°R; V — = ~ or ^ ^ 

Pzv 3 T 3 Pi v 2 Tz 


Vz T 2 


p 2 v 2 k = PiVi k or 


Pj. _ Ivz\ k 
Pi \v 2 ) 

Equating these values of p 2 /p 2 


= 118.5 psia 
= 59.25 psia 

[Eq. (1:6)] 
[Ex. 2:12 A] 




Thus, since Jc = 1.4 (from Example 2:12R), 
T 2 y/( k ~» n /geoxvd^-D 


Vi 


(T 2 

V2 \Tz 


) 


Pi = 


CTz 


6 ^ 560 ) 
(0.37) (560) 


= (6)(1.713) 2 - 5 = (6) (3.85) = 23.1 ft 3 


Vi 


23.1 


= 8.98 psia 


Point 4: T 4 = T R = 560°R 

i/(fc-i) 


P* 


IT A 1 
(0.37) (560) 


11.55 


/oeoxdd^-i) 

= 3 \560/ = ( 3 )( 3 - 85 ) = n - 53 ft 3 

= 17.96 psia 


(6) Wi = 16 + (0.46) (118.5) (3) = 179.5 Btu;w 2 = 16 + (0.46) (59.25) (6) = 179.5 Btu 
u 3 = IQ + (0.46)(8.98)(23.1) = 111.5 Btu; w 4 = 16 + (0.46)(17.96)(11.55) 

= 111.5 Btu 

Qs = u 2 - in + i W 2 /J = 179.5 - 179.5 + AV 2 /J = X W 2 /J 
For the isothermal processes 1-2 and 3-4, pv = const = CT or P = 144CT /v. 

iW 2 = PdV = 144 CTs jl‘ J = UlCTs log, ^ 

= (144)(0.37)(960) log, | = (51,100)(0.693) = 35,400 ft-lb 
Qs = iW 2 /J = 35,400/778 = 45.5 Btu 











THE REVERSIBLE PROCESS AND REVERSIBLE CYCLE 73 


Similarly, 


1 TF 4 = [J/PdV = 144 CTr f Vi - = 144C7 7 * log e - = (144) (0.37) (560) log e 

JV 3 JV 3 V Vi 


11.55 

23.1 


= (29,800)(-0.693) = -20,650 ft-lb 


Qr = — 


>w< 


-20,650 


= 26.5 Btu 


J 778 

Net work of cycle = J(Q S - Qr) = 778(45.5 - 26.5) = 14,750 ft-lb or 19 Btu 


(c) i W 2 = 35,400 ft-lb; 2 W 3 = J(u 2 - u 3 ) = 778(179.5 - 111.5) = 52,900 ft-lb 
3 W t = -20,650 ft-lb; i = J{ia - uQ = 778(111.5 - 179.5) = - 52,900 ft-lb 
Net work of cycle = y W 2 + 2 W 3 + 3 W 4 + iW i 

= 35,400 + 52,900 - 20,650 - 52,900 = 14,750 ft-lb 

14,750 


( d ) Ratio of net work of cycle to gross work of expansion = 


35,400 + 52,900 
= 0.167 


The low value of this ratio makes clear one of the reasons why the Carnot cycle is not a 
satisfactory cycle on which to base the operation of a real engine. The effects of 
friction on the real engine might be expected to be of such magnitude as to decrease 
the work of expansion by about 10 per cent and to increase the work of compression 
in about the same proportion. Applying this proportion to the example, the work of 
expansion would be 0.9(35,400 + 52,900) = 79,500 ft-lb and the work of compression 
1.1(— 20,650 — 52,900) = —81,000 ft-lb. Thus the real engine, instead of supplying 
work to other systems, would require work from those other systems to help in over¬ 
coming frictional effects. 


(e) The efficiency of the reversible Carnot engine is, for the conditions of this example, 
Qs-Qr 45.5 - 26.5 n 

V = -- = -- =* U.41/ 

Qs 45.5 

4:10. The Stirling Cycle and Stirling 
Engine. Many reversible cycles will 
suggest themselves to the reader; all 
that is necessary is to set up any com¬ 
bination of individually reversible 
processes that will eventually return 
the system to its original state. A 
few of these cycles may also have the 
important characteristic of the Carnot 
cycle in that they exchange heat Fig. 4:8. The Stirling cycle. 

only with two external reservoirs, each of which is at constant tempera¬ 
ture. An example of this is the Stirling cycle. 

The Stirling cycle is illustrated in Fig. 4:8. It consists of an isothermal 
1-2 at the constant temperature Ts of the source and an isothermal 3-4 at 
the constant temperature T R of the refrigerator. These are connected by 
the constant-volume processes 2-3 and 4-1. During the process 2-3 the 
temperature of the working system drops from T s to T R and, during 4-1, 
increases again from T R to T s . Since these two processes take place over 
the same temperature range, even though at different system volumes, 













74 


BASIC ENGINEERING THERMODYNAMICS 


regeneration, as described in Art. 4:6, may be employed and the heat that 
is discharged by the system during the process represented by the path 
2-3 may be restored to the system during process 4-1. These heat 
amounts would then not be included in Q R or Q s . This is equivalent to 
making the capacitors described in Art. 4:6 a part of the system. 

The operation of the Stirling engine is based on the Stirling cycle as 
described above. It is similar to the Carnot engine in that it is reversible, 
with all which that implies, when the source from which it draws heat and 
the refrigerator to which heat is discharged are at, respectively, constant 
temperature. It is dissimilar in the relative complexity of the apparatus 
which is required to carry out the necessary operations. Yet real Stirling 
engines are built and operated. These are not reversible engines, of 
course, and their efficiency falls far below that which we shall later calcu¬ 
late as applying to the reversible engine. The point of principal interest 
about the Stirling engine is that, as a reversible engine, it furnishes a basis 
of comparison with the reversible Carnot engine under equivalent condi¬ 
tions of operation, viz., constant temperature of source and constant 
refrigerator temperature. 

The features of the design of the Stirling engine that enable it to operate 
on the Stirling cycle will be discussed at a later and more appropriate 
stage in our discussion. For the present it need only be said that the 
Stirling engine requires two pistons, a working piston and a displacer 
piston. -The former moves during the isothermal processes, the latter as 
regeneration is employed to accompany the constant-volume processes. 

Example 4:10. The system contained within a Stirling engine consists of 1 lb of 
the gas for which equations of state are as in Example 1 : 10 . Referring to the nota¬ 
tion of Fig. 4:8, V x = 3 ft 3 , V 2 = 6 ft 3 , t s = 500°F, and t R = 100°F. (a) Calculate 

the pressure, volume, and temperature at points 1 , 2 , 3 , and 4 around the cycle. ( 6 ) 
Calculate the heat rejected to the regenerator ( 2 Q 3 ) and restored to the cycle during 
process 4-1. (c) Calculate Qs and Q R per cycle and, by difference, the net work of the 

cycle, (d) Compare the net work of the cycle with the gross work of expansion, 
(e) What is the efficiency of the reversible engine? 

Solution: 

(a) Point 1: 7\ = Ts = 960°R;t>, = V, = 3ft*;p, = — 1 = < 0 - 37 >( 960 ) = 118 . 5 ps i a 

Vi 3 

Point 2: T 2 — Ts = 960°R; v 2 — V 2 = 6 ft 3 ; p 2 = 59.25 psia 

Point 3: T, = T R = 560°R; v, = v t = 6 ft"; p, = — 1 = (°- 37 H 560 ) = 34 5 psia 

vz 6 

Point 4: Th = T R = 560°R; V\ = V\ = 3 ft 3 ; p± — -- = 69.0 psia 

( b ) ui = u 2 = 179.5 Btu [E x 4 . 9 ] 

u z = 16 + (0.46)(34.5)(6) = 111.5 Btu; ua = 16 + (0.46)(69)(3) = 111.5 Btu 

Heat rejected to regenerator = - 2 Q 3 = -(w 3 - u 2 ) = -(111.5 - 179.5) = 68 Btu 

Heat restored from regenerator = 4 £h = u x — u i - 179.5 - 111.5 = 68 Btu 







THE REVERSIBLE PROCESS AND REVERSIBLE CYCLE 75 


(c) Qs = iip = 45.5 Btu; Qr = 


iW 4 

J 


144CT D , Vi 
— J — R l °8 e X 

J Vz 


(144)(0.37)(560), 3 

- -778- ,0g ‘ 6 

= -(38.3)(-0.693) = 26.5 Btu 

Net work of cycle = J(Q S - Q R ) = 778(45.5 - 26.5) = 14,750 ft-lb or 19 Btu 


(d) Ratio of net work of cycle to gross work of expansion = 14,750/^2 = 14,750/ 
(778)(45.5) = 0.417. The higher value of this ratio as compared with the correspond¬ 
ing ratio for the equivalent Carnot cycle indicates that the real engine to operate on 
the Stirling cycle would not be handicapped in the same degree as the real Carnot 
engine, k or instance, if we apply the assumptions of Example 4:9, the work of expan¬ 
sion for this real Stirling engine would be (0.9) (778) (45.5) = 31,800 ft-lb and the work 
of compression (1.1)(778)(—26.5) = —22,700 ft-lb, leaving a net dividend of 9100 
ft-lb of useful work. On the other hand, the introduction of regeneration into the 
Stirling-engine cycle causes additional irreversibilities in the operation of the real 
Stirling engine with which the real Carnot engine would not have to contend. 

(e) The efficiency of the reversible Stirling engine is, for the conditions of this example, 


V 


Qs - Qr _ 45.5 - 26.5 
Qs 45.5 


0.417 


It will be observed that this is the same efficiency that was calculated to apply to 
the reversible Carnot engine which operated under equivalent conditions as to the 
temperatures of source and refrigerator. 

Problems 

1 . Show that, in reversible adiabatic steady flow through an open system, the 
relation between the pressure and the specific volume of an ideal gas may be expressed 
as Pv k = const. [Hint: Consider that a unit weight of the gas constitutes a second 
system (a closed system) with which the observer travels as it moves through the open 
system.] 

2. Work Example 4:8, assuming the gas is (a) gas W ; (6) gas X ; (c) gas Z. 

3. In solving Example 4:8, what assumption was made with respect to the veloci¬ 
ties at entrance to and exit from the open system contained within the turbine? If 
these velocities had been, respectively, 100 and 500 fps, how would the answers have 
been changed, if at all? 

4. A gas at 1200°F moves steadily through a reversible isothermal turbine. The 
pressure at entrance is 150 psig, and the exhaust is to atmospheric pressure. Assume 
that the velocities and elevations at entrance and exit are equal. If the gas is (a) 
gas W, ( b ) gas X, (c ) gas Y, ( d ) gas Z, calculate the work delivered to the turbine 
shaft and the rate of heat flow into the turbine per pound of gas flow. (Hint: The 
amount of work delivered to the turbine shaft per pound of gas flow is equal to the 
amount of work observed to accompany the expansion of a 1-lb closed system as it 
moves through the turbine by an observer who travels with the system minus the net 
flow work, PiVi — P 2 V 2 , which is expended in the course of entering and leaving the 
turbine rather than in turning the shaft.) 

5. Work Example 4:9, assuming the system consists of 1 lb of (a) gas W; (6) gas 
X ; (c) gas Z. Compare your answers with those obtained in the solution of the exam¬ 
ple for gas Y, and note any similarities. 

6 . (a) Work Example 4:9, changing ts to 1000°F. Repeat this solution using (6) 
gas W ; (c) gas X; (d) gas Z. What effect does this change have on the cycle efficiency? 








76 


BASIC ENGINEERING THERMODYNAMICS 


7. (a) Work Example 4:9, changing tR to 200°F. Repeat this solution, using (6) 
gas W ; (c) gas X ; (d) gas Z. Discuss the effect of this change on the cycle efficiency. 

8 . Work Example 4:10, assuming that the'system consists of 1 lb of (a) gas W; 

(b) gas X ; (c) gas Z. Compare your answers with those of the example, and state 
whether or not each gas is to be preferred to gas Y and why. 

9. Work Example 4:10, changing ts to 1000°F. 

10. Work Example 4:10, changing tR to 200°F. 

11 . If regeneration is not employed in the Stirling cycle of Example 4:10 but 
the state path is the same and all processes are maximum-work (reversible) pro¬ 
cesses, what is the efficiency of the cycle? How does the average temperature at 
which heat was received from the source compare for this cycle with the original 
(Stirling) cycle? What is the effect on the average temperature at which heat is 
rejected to the refrigerator? 

12. A reversible cycle encloses an area of 3.5 in. 2 when plotted on a pV diagram on 
which 1 in. denotes a pressure differential of 80 psi and a volume change of 0.5 ft 3 . 
When traversed clockwise, 50 Btu is rejected to the refrigerator for each traversal of 
the cycle. What is the cycle efficiency? 

13. Are the following processes reversible or irreversible? Neglect heat flow except 
as specifically stated to take place. If a process is reversible, show how all systems 
may be returned to their original states. If irreversible, classify the irreversibility 
as to cause or causes.. 

(а) A weight slides frictionlessly down an inclined plane. 

(б) A weight slides down a rough inclined plane; it is slowed by friction. 

(c) A perfectly elastic spring is elongated. 

( d ) A marble drops from the rim of a bowl and eventually comes to rest at the bot¬ 
tom of the bowl. 

( e ) A fluid expands slowly behind a frictionless piston in an insulated cylinder. 

(/) Water is confined under a piston which exerts a constant pressure on its surface; 
the temperature of the water is raised by stirring. 

(i g ) The process outlined in (/) continues until the water evaporates. 

( h ) Water confined as in (/) increases in temperature as the result of heat received 
from a source at constant temperature. 

(i) A perfectly elastic ball is dropped through a void, striking a rigid floor. 

(j) The ball of (i) drops through the air. 

(, k ) A gas is confined in an insulated cylinder beneath a weighted piston which is 
supported on a stop. The stop is removed and the piston drops suddenly until the 
pressure of the gas becomes sufficient to support its weight. 

(/) Steam flows through a frictionless nozzle, decreasing in pressure and increasing 
in velocity. 

(m) Steam flows through a small orifice from a chamber at high pressure to a cham¬ 
ber at considerably lower pressure. Eddies are formed below the orifice which are 
gradually damped out. 

(n) The steam jet leaving the nozzle in ( l ) is expelled into a large chamber in which 
eddies are formed and which it leaves at low velocity. 

(o) A stream of water at 80°F is mixed with a stream of steam at 212°F. The 
steam is condensed and the mixture leaves at 200°F. 

14. Each of the following processes is irreversible. With respect to the systems 
defined by the italicized words, state whether the irreversibility is external or internal. 
When internal, can you define a boundary which will separate the system into two 
systems which remain distinct from each other and for at least one of which there is no 
internal irreversibility? 


THE REVERSIBLE PROCESS AND REVERSIBLE CYCLE 77 


(а) A closed rigid container with insulated walls contains a gas and an electrical 
heating element to which current is supplied from a source external to the container. 
The temperature and pressure of the gas increase. 

(б) A gas is confined within a cylinder with nonconducting walls and head. The 
gas is compressed by the rapid inward movement of a frictionless piston. 

(c) A rigid container is divided into two compartments by a rigid partition. One 
compartment contains gas W at pressure P and temperature T; the other contains 
gas X at pressure P and temperature T. An opening is made in the partition and 
the two gases mix. 

(d) Same as (c) except that the pressures are equal but the original temperatures 
differ. 

(e) Same as (c) except that both the temperatures and the pressures are equal in 
the two compartments before the partition is ruptured. 

(/) A copjper hall at 50°F is placed in a pail of water at 120 °F. The ball and the 
water reach a common temperature. 

( g ) A closed rigid container filled with a gas is immersed in a bath of hot water. 
The temperature of the container and its contents rises. 

( h ) Two gallons of water at 100°F are mixed with 1 gal of water at 70 °F. A common 
temperature is reached. 

(i) A stream of steam flows through a reducing valve, decreasing in pressure but 
doing no external work. Downstream from the valve the elevation and the velocity 
are the same as upstream. 

Symbols 

C constant 
h specific enthalpy 
J proportionality factor 
k a constant ratio 
p pressure, psi 
P pressure; pressure, psf 
Q heat flow; rate of heat flow 
T absolute temperature 
u specific internal energy 
U internal energy of a system 
v specific volume 
V volume of a system 
W work 

Greek Letters 

r/ efficiency of a heat engine 

Subscripts 

R refrigerator 
S source 


CHAPTER 5 


THE SECOND LAW OF THERMODYNAMICS 

5:1. Inadequacy of the First Law. The First Law states that when 
energy in one form disappears energy in another form must make its 
appearance in equivalent amount. It does not attempt to establish 
conditions under which such energy transfers may take place or to limit 
the extent to which conversion may be effected. Thus, as based on the 
First Law, the work that would result from the adiabatic expansion of a 
closed system was shown to be equal to the decrease of internal energy of 
the system. No limit was set upon the amount of that decrease until we 
went beyond the scope of the First Law to specify that if the 'process was a 
maximum-work process the rate of change of internal energy with respect 
to system volume would equal the negative pressure. We were then, 
although not informed of it at the time, in the realm of the Second Law. 

Many other instances may be cited of observations that transcend the 
implications of the First Law. The elementary observation, for example, 
that the direction of spontaneous heat flow is always toward the system at 
lower temperature is an encroachment on Second Law territory. The 
whole concept of reversibility of which we treated in the last chapter is 
based on Second Law principles. A statement of the Second Law and an 
examination of its implications can no longer be delayed. 

5:2. The Second Law. Formal statements of the Second Law are 
based on observations of its operation. The observations that may be 
made are many and various as has been indicated above, and it is not 
surprising to find that a multitude of statements are proposed to express 
the meaning of the law. One of the earliest and simplest was offered by 
Rudolf Clausius, a nineteenth-century scientist; it follows: 

Heat cannot , of itself , pass from a colder to a hotter system. A second 
statement, credited to Planck, 1 is as follows: 

It is impossible to construct an engine that will work in a complete cycle 
and produce no effect except to raise a weight and exchange heat with a single 
reservoir. 

Planck’s version of the Second Law is well adapted to the purposes of 
engineering thermodynamics, and our later references to the Second Law 
will in general be based on his statement. 

1 Max Planck, “Treatise on Thermodynamics,” Longmans, Green & Co., New 
York, 1927. 


78 


THE SECOND LAW OF THERMODYNAMICS 


79 


The two statements of the Second Law that are presented above are 
equivalent each to the other, for it may be shown that Planck’s statement, 
if false, will cause the Second Law as stated by Clausius also to be false. 
Assume, contrary to Planck’s statement, that a machine is available that 
will continuously change the heat received from a single reservoir into 
work. By the First Law this conversion must be complete, and all of the 
heat received will be converted into work. As illustrated in Fig. 5:1, 
the work delivered by this machine (P) will be used to operate a Carnot 
heat pump (C) which will draw Q R heat units per unit of time from a low- 
temperature refrigerator and will discharge Q s heat units (larger than Q R 
by the amount of work required to drive the Carnot heat pump) to the 
reservoir at higher temperature T s that is the source of heat supply for P. 
This reservoir need supply only Qs — Qr heat units to P while it receives 
Qs heat units from C. The Second Law, as stated by Clausius, has been 
proved false if Planck’s statement is not true since no systems except the 
reservoirs at T R and T s have been affected and heat has passed from the 
one at lower to the one at higher tem¬ 
perature. 

The Second Law, like the First Law, 
may be proved only by the failure of all 
attempts to disprove it. If it were not 
true, all worry as to the continued avail¬ 
ability of fuels as a source of heat and 
power would vanish, for it would be pos¬ 
sible to tap the limitless stores of energy 
in the seas and the atmosphere for this 
purpose. In Fig. 5:1, if the rate at 
which engine P receives heat is increased 
to Qs units per unit of time, that engine 
will provide a surplus of power equal to 
Q r . In this case the high-temperature 
reservoir has given only as much heat as 
it received from the Carnot heat pump and may be replaced by a conduc¬ 
tor. No violation of the First Law has occurred since the work delivery 
was at the rate at which energy in the form of heat was supplied by the 
refrigerator. But the net work output is at the expense of the thermal 
energy of one reservoir only, and the Second Law is violated. 

Not only would fuel be saved, but there would be an additional gain of 
no small importance when a need existed to create a temperature below 
that of surrounding systems. For the space to be cooled could provide a 
source of heat and thereby have its temperature lowered. To provide a 
graphic example and carry this idea to its ultimate of absurdity, our homes 
could be air-conditioned in hot weather not only without expense for 



Fig. 5:1. The Second Law. 










80 


BASIC ENGINEERING THERMODYNAMICS 


power but even with a net dividend of power that could be used to operate 
the vacuum cleaner and the toaster. This example illustrates why a 
machine of this kind is often called a perpetual-motion machine of the 
second kind. The first class of perpetual-motion machine is a contradic¬ 
tion of the First Law since these machines are designed to create energy. 
This second class of perpetual-motion machine involves no violation of 
the First Law, as has been shown, but the idea behind it is just as absurd 
from a thermodynamic standpoint. The Second Law may be said to 
deny the feasibility of a perpetual-motion machine of the second kind. 

6:3. Reversibility and the Second Law. It is of interest to apply 
Planck’s statement of the Second Law to the concept of reversibility 
that was introduced in Chap. 4. In the course of that discussion we 
arrived at the conclusion that friction, unrestrained expansion, or the flow 
of heat across a finite temperature interval would always cause irreversi¬ 
bility in a process. It will be shown that if a process which involved any 
of these effects were reversible a machine of the type declared impossible 
by the Second Law would be a possible result. 

In Chap. 4 it has been reasoned that maximum work is associated with 
the reversible process. Let us first show that this reasoning is in accord 
with the Second Law. We shall assume that a process may be found 
which, with the system in communication with a single reservoir only, 
will deliver more work than a reversible process that takes place with the 
system in communication only with the same reservoir and that connects 
the same two end states. By definition of reversibility, a third process 
must also be possible which is the reverse of the reversible process and 
which will return the system to its original state, affecting only the single 
reservoir and requiring from that reservoir only the same amount of work 
that was transferred from it during the reversible process. But this 
amount of work is less than that delivered by the system during the 
assumed process, and a cycle may be formed of the assumed process and 
this third process for which J'dW > 0. Since the system was in com¬ 
munication with only a single reservoir, a contradiction of Second Law 
principles is observed and the assumption must be abandoned. 

It has been shown in Chap. 2 that friction and unrestrained expansion 
cause the work associated with the tracing of a given state path to be less 
than the maximum. Their presence during any process must, according 
to the preceding paragraph, make that process irreversible. 

It remains only to show that the flow of heat across a finite temperature 
interval is irreversible. Heat may flow from a reservoir at higher to a 
system at lower temperature without affecting other systems. But the 
possibility of heat flow in the opposite direction without affecting other 
systems is directly denied by Clausius’ statement of the Second Law, and 
his statement has been shown to be the equivalent of that of Planck. 


THE SECOND LAW OF THERMODYNAMICS 


81 


5.4. The Second Law and the Carnot Engine. Even a cursory exami¬ 
nation of the Second Law will indicate that one of its important implica¬ 
tions is to the el feet that, for the continuous conversion of heat into work 
(power), a minimum of two reservoirs must be available. A Carnot 
engine is a device that could be employed for the purpose. In order that 
the operation ol the Carnot engine may return a net dividend of power, it 
is required that its source and its refrigerator shall differ in their tempera¬ 
tures. T hus it is seen that the characteristic in which the two reservoirs 
required by the Second Law must differ is their temperature. 

Further, it has been shown that the heat discharged by a Carnot engine 
to its refrigerator will decrease with respect to a given amount of heat 
which it receives from a source at fixed temperature as the temperature 
of the refrigerator is lowered (see Art. 4:9) and that the net dividend of 
work that it delivers will correspondingly increase. Since this dividend 
of work can only equal the difference, Q s — Qr, between the heat received 
from the source and that rejected to the refrigerator, the implication of 
the Second Law that not all of the energy received as heat from the source 
can be continuously converted into work is sustained. 

The Carnot engine is shortly to become our tool for evaluating the 
effects of the operation of the Second Law on a quantitative basis. 
Before it can be used for that purpose, it becomes necessary to prove that 
its performance cannot be exceeded by that of any other engine which 
operates under equivalent conditions. 

5:5. The Carnot Principle. The Carnot engine derives its importance 
not because of its practical utility but because it is an example of a reversi¬ 
ble engine. It was devised by Carnot to accompany the statement of a 
principle of great importance in thermodynamics: 

No engine working continuously between given fixed and uniform tempera¬ 
tures of source and refrigerator can have a greater efficiency than a reversible 
engine that operates between these same reservoirs. 

The proof of the Carnot principle is based on the Second Law. Let it be 
assumed that the principle is not valid and that an engine may accordingly 
be constructed which, when operating between a given source and a given 
refrigerator, will have an efficiency higher than that of a Carnot (reversi¬ 
ble) engine which operates between the same reservoirs. Both engines 
will be placed between source and refrigerator as shown in Fig. 5:2. 
When operated as engines so that each withdraws heat from the source at 
a rate (per unit of time) which will be designated as Q s , the Carnot engine 
C will reject Q R units of heat and the irreversible engine I will discharge 
Qa units of heat to the refrigerator in the same time interval. By 
assumption the efficiency of engine I exceeds that of C, and therefore 
Qr > Qa- 

The Carnot engine C is reversible and therefore can and will be reversed. 


82 


BASIC ENGINEERING THERMODYNAMICS 


As a heat pump it will receive Qr units of heat from the refrigerator, dis¬ 
charge Q s units to the source, and require that work be supplied at the 
rate Qs — Qr to drive it. The work output of engine / is Q s — Qa, and 
this will be more than sufficient to operate engine C; a work dividend 
available for other purposes and equal to Q R — Qa will result. Examina¬ 
tion indicates that the source at T s may be replaced by a conductor (since 
it gives and receives heat at the same rate) and that the work dividend 
Qr — Qa is solely at the expense of an equivalent amount of energy 
removed as heat from the refrigerator. This is a violation of the Second 



Fig. 5:2. The Carnot principle. 


Law, and the assumption with which this analysis began must be rejected 
as impossible. The Carnot principle is therefore confirmed. 

Two important corollaries of the Carnot principle are stated below: 

1. All reversible engines , when operating between the same temperatures of 
source and refrigerator, will have the same efficiency. 

2. The efficiency of a reversible engine will depend only on the temperatures 
of the source and the refrigerator and will be independent of the working 
substance. 

The method of proof of each of these corollaries is similar in character 
to the proof of the Carnot principle; it is assumed that the opposite is 
true, and this assumption is then shown to be absurd. The detailed 
steps in each proof are left to the reader. 

The first corollary indicates that the reversible Stirling engine would 
have the same efficiency as the Carnot engine when operating under 
equivalent conditions and could therefore replace the Carnot as a device 
for evaluating Second Law effects. That the Carnot engine is the 
accepted tool for the purpose is due in part to the greater simplicity of 
the cycle on which it operates in that heat exchange takes place only with 













THE SECOND LAW OF THERMODYNAMICS 83 

source and reirigerator and there is no necessity for regeneration (see 
Art. 4:6). 

The second corollary leads directly to the concept of a thermodynamic 
scale of temperature to be discussed in the following article. 

6:6. The Thermodynamic Scale of Temperature. To Carnot is 
credited the discovery that the maximum thermal efficiency possible for a 
heat engine is that of a reversible engine and that this maximum efficiency 
is some function of the temperatures of source and refrigerator. In 
Carnot's time temperatures were based on arbitrarily chosen properties 
of the thermometer. Within the range of the thermometer the relative 
hotness of an object could be determined and compared with those of 
other objects but on only a qualitative basis. The temperatures of a 
number of objects furnished a means of arranging these in order of their 
hotness; beyond that, temperature had little significance except that tem¬ 
peratures were reproducible and that equality of temperature could there¬ 
fore be established. 

When he advanced his theory, Carnot had no knowledge of what func¬ 
tion of the temperatures of source and refrigerator represented the 
efficiency of the reversible engine. It remained for Lord Kelvin, a quar¬ 
ter of a century later, to realize the significance of Carnot's discoveries 
and to suggest that, based on the efficiency of the reversible engine, 
a temperature scale could be devised that would be independent of 
any property of the thermometric substance. The temperature scale 
which Kelvin proposed might well be called the thermodynamic scale of 
temperature. 

Suppose three reservoirs at temperatures h, t 2 , and t 3 , of which t\ is the 
hottest and t 3 the coldest as illustrated in Fig. 5:3. A reversible engine A 
will receive heat at a rate that will be designated as Q i from the reservoir 
at t\ (acting as its source) and discharge heat at the rate Q 2 to the reservoir 
at t 2 as its refrigerator. A second reversible engine B will receive heat at 
the rate Q 2 from the reservoir at t 2 (which thus acts as its source) and will 
discharge heat at the rate Q 3 to the reservoir at t 3 as its refrigerator. By 
the First Law, engine A will deliver work at a rate equal to the difference 
between Qi and Q 2 , or Qi — Q 2 , and engine B will have a rate of work 
delivery equal to Q 2 — Q 3 . 

Reversible engine C is placed between the reservoir at ti and the reser¬ 
voir at t 3 , the first acting as its source and the second as its refrigerator. 
It will be operated simultaneously with engines A and B and at a rate 
such that Qi units of heat are drawn from the source in the same time 
that this reservoir supplies the same amount of heat to engine A. The 
rate at which it rejects heat to its refrigerator, the reservoir at t 3 , will 
be designated as Q 3 , and its work output is therefore at the rate Qi — Q 3 . 

For Q i units of heat removed from the reservoir at t h the sum of the 


84 


BASIC ENGINEERING THERMODYNAMICS 


work output of engines A and B is Qi — Qz, and the work output oi 
engine C is Qi — Qz. If W A + W B > Wc, engine C may be reversed and 
driven as a heat pump by means of power furnished by engines A and B, 
leaving a surplus of power for other purposes. A violation of the Second 
Law will be the result since the reservoirs at h and t 2 will both be acting 
merely as conductors, there being no net heat flow to or from them, and 
the surplus of power mentioned above is solely at the expense of heat 



Fig. 5:3. The concept of a thermodynamic scale of temperature. 

taken from the reservoir at U. If, on the other hand, Wc > W A + W B , 
the reversal of engines A and B to act as heat pumps will produce the 
same result. Therefore W A + W B = Wc, Qi ~ Qs = Qi — Q3, and 
Q3 == Q3. 

Equation (2:15) may be expressed as 

,. ■ - 1 <*■> 

As it applies to the conditions illustrated in Fig. 5:3, this equation would 
become 17 = 1 — Qs/Qi and this would be the efficiency of both engine C 
and the combination of engines A and B. As the temperature of the 
reservoir at U was decreased, the efficiency would increase and Qz would 
correspondingly decrease. This could be used as the basis of a tempera¬ 
ture scale on which U could be rated with respect to t\ in the proportion 
that Qz bears to Q 1 . Let us temporarily designate temperatures on this 

















THE SECOND LAW OF THERMODYNAMICS 


85 


proposed thermodynamic scale of temperature as 0 h 0 2 , 0 3 , etc. Then 


Q 3 _ Qk _ 03 _ 0R 

Qi Qs 0i 0s 


and 


_ 1 0R 

^?rev engine J- ~Z~ 

VS 


0S 0fl 

d s 


(5:2) 


The Kelvin and Rankine scales of temperature were devised to meet 
the requirements of this thermodynamic scale of temperature, and they 
vary so slightly from it that T on either scale may be substituted for 0 in 
Eq. (5:2). The efficiency of a reversible engine when operating between 
two reservoirs may then be stated as 


7/rev 


engine 


T r _ T s - T r 

T s ”” T s 


(5:3) 


Naturally, the same scale of absolute temperature must be used to express 
both T s and T R when substitution is made in Eq. (5:3). 


Example 5:6. Using Eq. (5:3), calculate the efficiency of the Carnot and Stirling 
engines of Examples 4:9 and 4:10. 

Solution. For both engines, 


V 


T s - T r 960 - 560 
T s 960 


0.417 


This answer agrees with the result of the calculation of the efficiency as carried out in 
the two examples. 


As the efficiency of the reversible engine increased toward unity as a 
limit, Q r would correspondingly decrease until the heat discharged to the 
refrigerator became zero. This would also constitute a zero point on the 
thermodynamic scale of temperature. As the temperature of the refrig¬ 
erator declined below this point and became negative, it would be noted 
that the heat rejected to the refrigerator would have become a heat supply 
from that reservoir. This energy and that which the engine received 
from its source could be replaced by heat flow from a single reservoir at a 
temperature that was sufficiently high, and a violation of the Second Law 
would be apparent. It is impossible to attain negative temperatures on the 
thermodynamic scale of temperature. The corresponding temperature on 
the Kelvin and Rankine scales is therefore designated as the absolute zero 
of temperature. 

The lowest temperature that develops as the result of natural processes 
on the earth’s surface is about 200°K. In order to demonstrate tempera¬ 
tures which approach absolute zero it is therefore necessary to employ 
some kind of “heat pump” to draw heat from the body to be cooled, as 
the refrigerator, and discharge it to a reservoir at some higher level of 
temperature. When, as the result of the removal of this heat, the tem¬ 
perature of the cold body approaches absolute zero, the ratio of the work 
required to drive the heat pump to the heat withdrawn from the cold body 







86 


BASIC ENGINEERING THERMODYNAMICS 


approaches infinity. Unless the cold body may be perfectly insulated, 
which is never possible in practice, it is in the meantime receiving heat at 
a finite rate from surrounding systems. In the absence of a perfect insula¬ 
tor , absolute zero of temperature cannot be attained. 

Problems 

1. For any of the nonflow processes outlined in Prob. 13, Chap. 4, which you decide 
are irreversible, show that a perpetual-motion machine of the second kind would result 
if the process were reversible. 

2. Show, for each of the irreversible processes described in Prob. 14, Chap. 4, 
that a perpetual-motion machine of the second kind would be required to make it 
reversible. 

3. In Example 2:10 it was shown that it was possible to change all of the heat 
received by a system from a source at constant temperature into work as the result 
of an isothermal process. Why is this not a violation of the Second Law? 

4. A system consisting of an electric motor and connected storage battery can deliver 
work to surrounding systems and, at the same time, give off heat to the surroundings. 
Why is this not a violation of the Second Law? 

5. Prove the first corollary of the Carnot principle. 

6. Prove the second corollary of the Carnot principle. 

7. Prove that a system which is not receiving wmrk cannot convey heat from a cold 
to a hot body continuously. Describe a system which might do so over a limited 
period of time. 

8. Two Carnot engines A and B both receive heat from the same source at an 
unknown thermodynamic temperature which will be designated as 0 x . Both receive 
heat at the same rate Qi, but the magnitude of Q i is not known. Carnot engine A 
discharges heat to a sink of unknown temperature 0 2 and delivers a known amount of 
power. Engine B employs a sink of unknown temperature 0 3 and delivers half the 
power of engine A. Based only on the information as stated above, which of the 
following statements are necessarily true? (a) The efficiency of engine A is twice 
that of engine B. ( b ) 0 3 = 20 2 . (c) 0 2 = 20 3 . (d) 0i — 0 2 = 2(0i — 0 3 ). ( e ) The 
ratio 0 3 /0 2 may be computed. (/) The efficiency of engine A may be computed. 

9. Same as Prob. 8 except that Q i is known in magnitude. 

10. What is the maximum possible efficiency of an engine that receives heat from a 
source at 1000°F and discharges heat to a sink at 70°F? 

11. A Carnot engine delivers 50 hp while receiving heat from a source at 800°F and 
discharging heat to the atmosphere at 60°F. At what rate, in Btu per minute, does it 
receive heat? What is its efficiency? 

12. A reversible engine which does not operate on the Carnot cycle receives 100 Btu 
of heat per second from a constant-temperature source at 600°F and delivers 70 hp. 
If the temperature of its refrigerator is constant, what is that temperature? 

13. A Carnot engine that uses gas W as its working substance operates between 
source and sink temperatures of 1200 and 100°F. What is its efficiency? How would 
your answer have been affected if the working substance had been gas F? If it 
had been a nonideal gas? If it had been steam? 

14. A Carnot engine has an efficiency of 40 per cent and rejects heat at the rate of 
12,000 Btu/hr to a refrigerator at 70°F. What is the temperature of its source? 
What horsepower does the engine deliver? 

15. A reversible engine discharges 200 Btu /min to a refrigerator at a constant tem¬ 
perature of 40°F while delivering 5 hp. What is the (constant) temperature of its 
source? What is its efficiency? 


THE SECOND LAW OF THERMODYNAMICS 


87 


16. A Carnot engine uses an ideal gas as its working substance. At the end of the 
isothermal expansion, the pressure of the gas is 100 psia, and the volume of the con¬ 
tained system is 4 ft 3 . At the end of the adiabatic expansion, the pressure is 20 psia, 
and the volume is 12.6 ft 3 . What is the efficiency of the engine? 

17. At the beginning of the isothermal expansion, the pressure and volume of the 
ideal-gas system that is the working substance of a Carnot engine are 50 psia and 1 
ft 3 , respectively. At the end of the adiabatic expansion, the pressure is 15 psia, and 
the volume 2.5 ft 3 . What is the efficiency of the engine? 

18. A reversed Carnot engine removes 12,000 Btu/hr from a refrigerator at 10°F 
and discharges heat to the atmosphere at 90°F. What horsepower is required to 
drive it? 

19. A building that requires 100,000 Btu/hr to maintain its temperature at 70°F 
when the outside temperature is 20°F is to be heated by means of a reversed Carnot 
engine. What horsepower will be required to drive the engine? At what rate is heat 
removed from the outside air? 

Symbols 

Q rate of heat flow, per cycle or per unit time 
t scalar temperature 
T absolute temperature 

W rate of work delivery, per cycle or per unit time 
Greek Letters 

t) efficiency of a heat engine 
0 temperature on the thermodynamic scale 

Subscripts 

R refrigerator 
S source 


CHAPTER 6 


ENTROPY, A PROPERTY OF THE SYSTEM 

6:1. Entropy. It has been observed that it is possible to represent the 
work which accompanies a reversible process (a maximum-work process) 
on a diagram of which the coordinates are the pressure and the volume, 
both being properties of the system. A diagram that would perform the 
same function for the heat flow which takes place during a reversible 
process would be of equal or even greater usefulness. With reference to 
the maximum-work diagram, pressure is an intensive, volume an exten¬ 
sive property. Pressure may be regarded as that property which furnishes 
the incentive, or potential, for the system to do work. Differentials of 
pressure, infinitesimal for the reversible process, determine the direction 
of work flow. The property with respect to heat flow which is analogous 
to absolute pressure on the work diagram is temperature since differences 
in temperature determine the direction of heat flow. Absolute tempera¬ 
ture therefore will be used as the ordinate on the diagram for which we are 
searching. 

The second dimension, which must be a property of the system since a 
plotted point on the diagram must definitely establish a unique state of 
the simple system, is more difficult of selection. This dimension may, 
however, be determined in terms of the quantities by which it must be 
measured. We may rewrite Eq. (2:6) in differential form as dW = P dV. 
Equation (2:6) is valid only for a reversible process and, transposing, 

dV = (reversible) (6:1) 

Following the analogy which is proposed above, if we designate the sought- 
after property by the symbol S, then 

dS = ^ (reversible) (6:2) 

Equation (6:2) defines a quantity which we shall call entropy (en'tropy). 
If entropy can be shown to be a property of the system, the temperature- 
entropy diagram will have the same usefulness with respect to the meas¬ 
urement of reversible (maximum) heat flow that the pressure-volume 
diagram possesses in connection with the calculation of the maximum 
work which may accompany a given path. 

It remains to be proved that entropy is a property of the system. One 

88 



ENTROPY, A PROPERTY OF THE SYSTEM 


89 


of the methods of proof of the status of a quantity as a property of the 
system that was outlined in Chap. 1 consisted in showing that its line 
integral around any cycle was zero. If it can be shown that the line 
integral of dQ/T (reversible) is always zero, then the line integral of dS 
must be zero around any cycle and entropy will assume the status of a 
property of the system. The proof that the line integral of dQ/T is zero 
around any reversible cycle is equivalent to the requirement as set forth 
above, and this proof will be the next step in our discussion. 

Temperature and volume have approved status as independent proper¬ 
ties of the system. We may, as in Fig. 6:1, represent any reversible cycle 
on these coordinates by the closed 
curve R of that figure. The area en¬ 
closed within R will be divided into 
curved strips by the lines a, which rep¬ 
resent the paths followed during revers¬ 
ible adiabatic processes. Each pair of 
adjacent adiabatics may then be joined 
at the top and the bottom by horizontal 
lines i, which will intersect the curve R 
as shown. These horizontal lines rep¬ 
resent reversible isothermals and, to¬ 
gether with the reversible adiabatic 
lines which they connect, form individ¬ 
ual Carnot cycles into which the entire area enclosed within R is divided. 
As the distance between the lines a is reduced, the contour of the outline 
of these Carnot cycles will approach the closed curve R. 

For each of the individual Carnot cycles it has been shown in Art. 5:6 
that Qs/Ts = Qr/Tr [see Eq. (5:2)]. Since the heat transfers during the 
reversible isothermal processes take place at constant temperature, 
Q/T may replace /dQ/T (reversible). But Q R represents a rejection 
of heat by the working substance and, taking the sign of the heat flow into 
consideration, 



Fig. 6:1. Entropy as a property. 


Qs , Q 


Ti 


+ 


R 


= 0 


R 


( 6 : 3 ) 


In other words, the increase of the quantity along the upper isothermal is 
equaled by its decrease along the lower. The curved sides of each of 
these small cycles are reversible adiabatics during which there is no heat 
flow and for which /dQ/T is zero. The total change of the quantity 
around each of these segmentary cycles is therefore zero. But these 
cycles, taken together, duplicate the reversible cycle R as a limit, and we 
obtain 


^ (reversible) = 0 


(6:4) 







90 


BASIC ENGINEERING THERMODYNAMICS 


Example 6:1 A. Calculate the total change of entropy around the Carnot cycle 
of Example 4:9. 

Solution: 


For the reversible process 1-2, S 2 — Si = 

For the reversible process 2-3, S 3 — S 2 = 

For the reversible process 3-4, Si — S 3 = 

For the reversible process 4-1, Si — Si = 



Qs _ 45.5 
T s ~ 960 



0.0473 


—Q r __ —26.5 
T r 560 



-0.0473 


Around the reversible cycle, (j) ^ = (j) dS = 0.0473 + 0 + (—0.0473) + 0 —0 


The basic criterion on which the status of a quantity as a property is 
established is that its change between two given states of the system is 
independent of the path or process which is traversed by the system dur¬ 
ing that change of state. It may be shown that this requirement is satis¬ 
fied for entropy if Eq. (6:4) is valid. Let two reversible cycles l-m-2-n 
and l-ra-2-p, respectively, be plotted on a diagram having as its coordi¬ 
nates any two independent properties of the system P 1 and P 2 , as illus¬ 
trated in Fig. 6:2. These cycles have a common reversible process l-ra-2 

for which f ^ (reversible) is evidently the same for both cycles. But, 

based on Eq. (6:4), the line integral of 
dQ/T around either of these cycles 
must be zero, and we may therefore 
conclude that the total change of the 
quantity dQ/T between the states rep¬ 
resented by points 2 and 1 will be the 
same for any reversible process that may 
connect them. For the measurement 
of the change of entropy between the 
two states a reversible process must be 
selected if that change of entropy is to 
be evaluated as /dQ/T. For it is only 
in connection with the reversible proc¬ 
ess that the definition of entropy which is implied in Eq. (6:2) is valid. 
However, having established the amount by which the entropy has 
changed by applying Eq. (6:2) to any reversible process that connects the 
two states, this change of entropy will apply whenever the system 
changes from one state to the other even if the process that connects them 
is irreversible. 

Let us illustrate this important point more forcibly by returning to our 
analogy in which the proposed diagram for the measurement of heat flow 



Fig. 6:2. Entropy as a property. 








ENTROPY , A PROPERTY OF THE SYSTEM 


91 


during a reversible process was compared with the pressure-volume dia¬ 
gram and its use in measuring work transfer during the reversible (maxi¬ 
mum-work) process. Referring to Fig. 6:3, let the full lines m and n 
between points 1 and 2 represent the paths followed during any two 
reversible processes that connect these states of the system. The dashed 
line i will represent the path traversed during a third, but irreversible, 
process which has its inception at state 1 and ultimately brings the system 
to state 2. It has been shown that the maximum work that may accom¬ 
pany a given path will be developed when that path is carried out reversi¬ 
bly and that the amount of this work is represented by the area under the 
path on a pressure-volume diagram. Thus the strip areas dW m and dW n 
represent the work done by the system as segments of the reversible paths 
m and n, respectively, are traversed. Equation (6:1) may be applied to 
find the elementary change of volume p 
that took place, and the integration of 
these infinitesimal volume changes over 
the entire process would result in the to¬ 
tal change of volume during the process, 
or V 2 — V\. Further, no matter which 
of these reversible processes was used, 
the total change of volume would be the 
same. Now let us consider the same 
method as applied to the irreversible 
process i. A strip area (shown with¬ 
in dashed lines) may again be associ¬ 
ated with a segment of this path, and, 
following the same procedure as for the reversible processes m and n, the 


£ 



|V \5* 


N 










IV— 

. _ - 


i 1 



j | k 

^ 1 1 IT 

< 


1 ^ 

ii 

_u_ 



Fig. 6:3. Evaluation of change in 
volume. 


change of volume may be evaluated as F 2 — Fi = 


dAi 


But in this 


case the area of the strip does not represent the work that accompanied 
the elementary change of volume. Instead we know that this work is less 
than the segmentary area since the process is irreversible and therefore 
dWi < dAi. Returning to Eq. (6:1), we have seen that it is correct to 

write dV = dAi/P and V 2 — Fi = / tt’ Since dWi < dAi f it 


P 


is 


incorrect to substitute dWi for dAi in these equations, but inequalities 
may be written, as follows: 


dV > 


dW 


(irreversible) and V 2 — Fi > 


d W 
P 


(irreversible) 

(6:5) 


To include both reversible and irreversible processes in their scope, Eqs. 














92 


BASIC ENGINEERING THERMODYNAMICS 


(6:5) could be written 


dV > 


dW 

P 


and 



Vi > 



dW 

p 


( 6 : 6 ) 


In this form they would apply to any process. The equality would, of 
course, apply to any reversible process [see Eq. (6:1)]. 

To conclude our application of the analogy, note that the same total 
change of volume was obtained by applying Eq. (6:1) over any reversible 
process connecting states 1 and 2 of the system and that this change of 
volume, though not in that case obtainable as the integral of dW/P, also 
applied to any irreversible process connecting those states. Just as the 
work is a maximum for a reversible process, so is the heat flow also a 
maximum (see Chap. 2), and we may write expressions for the change of 
entropy analogous to Eqs. (6:6), 

dS> ^ and S, - Si > f ~ (6:7) 


The equality again applies to the reversible process. The inequality is in 
the algebraic sense that has previously been applied in the definition of 
maximum work. 

Summarizing, since entropy is a property, its change will be independent 
of the process by which a change of state is effected and will depend only on 
the end states. The integral of dQ/T may be used to evaluate the change of 
entropy only in a reversible process connecting the given end states. 

Example 6:15. The rate of heat flow to a system per unit increase of temperature 
during a certain reversible process is constant and equal to 0.7 Btu/°F. (a) Calculate 
the change of entropy during the process if the initial temperature is 100°F, the final 
temperature 300°F. ( b ) A second process which follows the same path between the 

same end states is accompanied by heat flow at a rate half as great as in part a. What 
is the change of entropy for this second process? (c) A third process connects the 
same end states over an entirely different path. What is the change of entropy? 

Solution: 

(a) = 0.7 or dQ = 0.7 dt = 0.7 dT 

S 2 - Si = (reversible) =0.7 ^ = °- 7 l°g e ^ = (0.7)(0.306) = 0.214 

( b ) The second process is evidently not reversible since the heat flow is less than the 
maximum that could accompany the given path. Thus the change of entropy is not 
obtainable as the integral of dQ/T. But the process connects the same end states as 
in part a, and since entropy is a property, the change of entropy will be the same, 
or S 2 — Si = 0.214. 


(c) No matter what path or process connected these two end states, the change of 
entropy, a property of the system, would be the same. Therefore, S* — Si = 0.214. 






ENTROPY, A PROPERTY OF THE SYSTEM 


93 


In evaluating changes of entropy it will be remembered that dQ = dU 
+ P dV/J for the simple closed system during a reversible process [see 
Eqs. (2:5) and (2:6)], and Eq. (6:2) may therefore be changed to the form 


dS 


dU PdV 
T JT 


( 6 : 8 ) 


This equation indicates that entropy is an extensive property since both 
the change of internal energy and the change of volume of the system with 
respect to a given change in its pressure and temperature will depend upon 
the mass of the system. The lower-case letter s will be used to designate 
specific entropy, or entropy per pound. The unit of entropy, since it is 
the quotient obtained by dividing the heat flow (Btu) by the absolute 
temperature, may be stated as Btu per degree Rankine if we employ the 
system of units used in this text. This unit of entropy is sometimes 
called the “rank.” It is customary in technical literature to omit any 
designation of the unit of entropy that is employed and allow it to be 
inferred from the context. Thus another unit of entropy might be the 
calorie per degree Kelvin. 

Example 6:1C. One pound of a substance having an equation of state as in Exam¬ 
ple 1:10 changes from an initial state at which p i = 30 psia, V\ = 10 ft 3 , to a final 



pressure and final volume of 20 psia and 14 ft 3 , respectively. Calculate the difference 
of entropy between the two states by applying Eq. (6:8). 

Solution. The required difference of entropy may be calculated as the integral of 
dU/T + P dV/JT over any path that connects these states. Two paths will be 
selected for the purpose, the change of entropy over each will be calculated, and the 
results will be compared. The first path will proceed from state 1 at constant temper¬ 
ature until the volume corresponding to state 2 is reached and thence at constant 
volume to state 2. This path is shown as path l-a-2 in the figure that accompanies 
this example. Then 


T a 

Pa 


= Ti = 


(30) (10) 


0.37 
(0.37)(811) 
14 


= 811°R; 
21.4 psia 


T 2 


(20)(14) 
0.37 


= 757°R; v a 


y 2 = 14 ft 3 









94 


BASIC ENGINEERING THERMODYNAMICS 


For segment 1-a of this path, from Eqs. (1:6) and (1:7), 

du = 0.46 d(pv) = (0.46)(0.37) dT = 0.17 dT = 0; p = 

P dv 144p dv _ (144) (300) dv _ dv 

JT 778 (778)(811)v ' ‘ v 


CT (0.37)(811) 300 


>. - ••=// t +/; ■ 11 + 00684 1 : f - °-° 684 ^ m - 00231 


T ' Ji JT 
For segment a-2, du = 0.17 dT and P dv = 0, 


s 2 

S2 




7^7 

171og e ^ = -0.0117 


811 


Si 


= (s a - Si) + (s 2 - Sa) = 0.0231 - 0.0117 = 0.0114 


Let us now select a second path which will proceed from state 1 at constant volume 
to state 6 at which the pressure is the same as at state 2 and thence at constant pres¬ 
sure until state 2 is reached. This path is also shown on the figure. Then, 

on • in fi.1 rp (20) (10) riiop 

Vb = P 2 = 20 psia; v b = Vi = 10 ft 3 ; T b = —Q37 - = 541 ^ 

For segment 16, du = 0.17 dT and P dv = 0 as shown above. 
s 6 - si = 0.17 log. Mi = -0.0690 

For segment 6-2, du = 0.17 dT. The pressure is constant and therefore, from Eq. 
(1:6), pdv = CdT = 0.37 dT, 


P dv 
JT 


(144) (0.37 )dT_ dT 


778 T 


T 


f 2 du p Pdv f 

J>7 + j> -JT- 017 in ~T + 00684 u 

= 0.2385 log. Hr = 0.0804 

§2 — Si = (sb — Si) T (S 2 — s&) = —0.0690 -t - 0.0804 = 0.0114 


Ti dT f T* dT 

n ~T - 0 2385 /n T 


This result agrees with that obtained by the calculation which was based on the path 
l-a-2. 


In utilizing this new property we shall differ in one respect from the 
methods used in connection with the analogous coordinate (volume) on 
the pressure-volume diagram. Volume may be, and usually is, measured 
in absolute terms although in connection with paths, processes, and cycles 
it is changes of volume that are of principal interest. The Third Law of 
thermodynamics states that the entropy of a system is zero when the tem¬ 
perature of the system is the absolute zero of temperature, and it is conceivable 
that the absolute value of the entropy corresponding to a given state of the 
system could be calculated by integrating dQ/T over some reversible path 
connecting a state at absolute zero to the given state. Inadequacy of 
available data at low levels of temperature makes this procedure impracti¬ 
cal, and it is customary either to content ourselves with a calculation of 
the change of entropy between states in which we are interested or, for the 
purpose of tabulating the values of entropy corresponding to various 
















ENTROPY , A PROPERTY OF THE SYSTEM 


95 


states of the system, to select some convenient reference point at which 
the entropy is arbitrarily given a value of zero. This completely serves 
our purpose since it is the changes of entropy with which we shall be con¬ 
cerned in practical thermodynamic computations rather than its absolute 
values. Note also that in the first steps of the procedure that is suggested 
above for the calculation of absolute entropy the change of entropy would 
be disproportionately large because of the low value of the denominator 
of dQ/T. 

Entropy is an abstract quantity. It cannot be observed as volume is 
observed and tangibly evaluated. Let us now see whether or not the 
effort that has been expended on the development of this new property 
promises to be rewarding. 

6:2. The Temperature-Entropy Diagram and the Reversible Process. 

Figure 6:4 shows a diagram for which the ordinate is the absolute tem¬ 
perature and the abscissa is the entropy. The path 1-2 represents a 
reversible process which is executed by a given simple closed system. The 
segmental area under this path which is shown in the figure has a height T 
and a width dS. From Eq. (6:2) its 
area is dQ, an element of heat flow. 

When dS is positive, as in moving to 
the right along the reversible path 1 - 2 , 
this heat flow is positive and repre¬ 
sents a passage of heat into the sys¬ 
tem; the opposite is true when the 
change of entropy is negative. 

Because a simple system is involved 
and entropy now has the status of a 
property, the path 1-2 represents a 
succession of states of the system and 
that path could be transferred to a 
pressure-volume chart since any two 
independent properties of the simple system (in this case the temperature 
and the entropy) will fix the values of all other properties. Also it has 
been shown previously that, if the process is reversible, the path must con¬ 
nect equilibrium states . 1 

The total area under the path (area 1-2-b-a), being the summation of 
elementary areas similar to that illustrated in the figure, represents the 
total heat flow that accompanies reversible process 1 - 2 . Conversely, the 

1 It is emphasized that this discussion applies only to the case when the process is 
reversible. If irreversible, the inequality [see Eq. (6:7)] will apply and dQ < 7 dS. 
For the irreversible process the path will be impossible to trace it it does not connect 
equilibrium states, but points may be established, as discussed in Art. 2.11, by stop¬ 
ping the process at intervals and allowing the system to assume an equilibrium state. 



Fig. 6:4. The TS diagram and the 
reversible process. 











96 


BASIC ENGINEERING THERMODYNAMICS 


change of entropy that takes place as the path is reversibly traversed can 
be found, by applying Eq. (6:2), to be [see Eqs. (6:7) and (6:8)] 

S 2 -S 1 = j* dS = j* ^ (reversible) = j' — + J* (6:9) 

Let us apply Eq. (6:9) to the problem of locating certain of the reversible 
processes discussed in Chap. 2 as corresponding paths on the temperature- 
entropy diagram. 

Example 6:2 A. During a reversible process that is represented by a straight line on 
a temperature-entropy diagram, the temperature of a system increases from 40 to 
200°F as its entropy decreases by 0.1 rank. What heat flow accompanies the process? 

Solution. The process is stated to be reversible, and the area under the path on the 
TS diagram therefore will measure the required heat flow. This area is trapezoidal, 
and thus 

1 Q 2 = T} -+ 12 AS = ( 5Q --^ 66 ° ) (-0.1) = -58 Btu 

It will be observed that the change of entropy supplies the same criterion for the 
determination of the direction of heat flow during a reversible process as the change of 
volume furnishes for the sign of work performed. For just as a continuous increase 
of volume throughout a reversible process indicates that the work of the process is 
positive, so a continuous increase of entropy indicates positive heat flow provided 
that the process is reversible. 


The Reversible Isothermal. The reversible-isothermal, or constant- 
temperature, path would of course appear as a horizontal line on the TS 

diagram. It is represented by path 
1-2 in Fig. 6:5. The heat flow that 
accompanies this process is easily 
measurable as the rectangular area 
1-2-6-a or as T(S 2 — Si). Here an 
advantage of the TS diagram as com¬ 
pared with the PV diagram makes its 
appearance. If it is necessary to cal¬ 
culate the heat flow that takes place 
during a reversible isothermal process 
and the PV diagram is our only tool 
for the purpose, the procedure as out¬ 
lined in Chap. 2 consists of three steps, 
as follows: 


/ 


2 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

/Q 2 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 a 

1 


1 

1 

1 

; 6 


S/ S 2 S 

Fig. 6:5. The TS diagram and the 
isothermal process. 


1. Measure the area under the path onaPF diagram, and so obtain the 
work that accompanies the process. 


2. Calculate the change of internal energy that takes place between 
states 1 and 2 by applying an appropriate equation of state. 

3. Substitute the values obtained by carrying out the first two steps in 












ENTROPY, A PROPERTY OF THE SYSTEM 


97 


Eq. (2:3), and thereby obtain the heat flow that is associated with the 
process. The temperature-entropy diagram, as a new tool at our com¬ 
mand, makes it possible to obtain the heat flow directly as the result of a 
single simple calculation provided that the entropies at the beginning and 
the end of the process are known. 

The Reversible Adiabatic. By definition, no heat flow takes place during 
an adiabatic process, and if the process is also reversible, there will be no 
change of entropy. 1 The process is illustrated in Fig. 6:6, path 1-2 being 
that followed during a reversible adiabatic which is accompanied by a 
decrease of temperature. Consistent 
with the definition of an adiabatic, the 
area under the path is observed to be 
zero. Again an example of the special 
usefulness of the temperature-entropy 
diagram is illustrated. In order to plot 
successive equilibrium states along a 
reversible adiabatic path on the pres¬ 
sure-volume diagram, it was first nec¬ 
essary to develop a relationship be¬ 
tween the pressure and the volume of 
the system that would apply along that 
path (see Example 2:12A) and then to 
use the resulting equation to plot the 
path. On the temperature-entropy diagram, on the other hand, the 
position of all points along the path is fixed by the requirement that the 
entropy shall not change and the path becomes simply a vertical line. 

The Reversible Constant-volume and Constant-pressure Paths. It has 
been shown that, in any reversible process, the change of entropy may be 
evaluated as the integral of dQ/T. For a constant-volume reversible 
process dQ may be replaced by Mc v dT (see Art. 2:7), and if, for con¬ 
venience, c v may be assumed to have a constant value over the tempera¬ 
ture range Ti to T 2 , then the corresponding change of entropy is 



S/ = S 2 

Fig. 6:6. The TS diagram and the 
reversible adiabatic process. 


S 2 — Si (constant volume) = Mc v 



dT 

T 


= Mc v log 


& e 


Ti 


( 6 : 10 ) 


The logarithmic curve that represents Eq. (6:10) is shown in Fig. 6:7 as 
the curve 1-2. The area l-2-b-a under this curve is the heat flow which 
accompanies this constant-volume reversible process. But no work is 
performed during a constant-volume reversible process, and, referring to 


1 Note that an irreversible adiabatic process could be accompanied by a change of 
entropy. Equation (6:7) indicates that the entropy would increase during that type 

of process. 









98 


BASIC ENGINEERING THERMODYNAMICS 


Eq. (2:3), this heat flow (and consequently the area 1-2 -b-a) is seen to be 
equal to the change of internal energy that takes place as a result of the 

process. This is indicated in Fig. 
6:7 by labeling the corresponding 
area I/ 2 — U i. Moreover, this 
area would have the same signifi¬ 
cance in this respect even if path 
1-2 had not been specified as revers¬ 
ibly performed, for the change of a 
property (in this case the internal 
energy) is dependent only on the 
end states and will be the same for 
any path or process that connects 
the same end states. Thus when 
the initial and final volumes of the 
system are equal, even if there was 
a change of volume in the course of 
the process, the change of internal energy will be equal to the area under 
a path on the TS diagram which passes through states having equal vol¬ 
umes and connects the given initial and final states. 

Proceeding in a similar manner, it may be shown that, with reference to 
the constant-pressure reversible process of Fig. 6:7, 



Fig. 6:7. The TS diagram and the 
constant-volume and constant-pressure 
process. 


S$ — Si (constant pressure) = Mc p 



= Mc p loge 



( 6 : 11 ) 


During any constant-pressure process an increase in the internal energy 
of the system will accompany an increase in its volume, and vice versa. 
Since the heat flow during the reversible process must account not only 
for the change of internal energy but also for the work that results from 
the changing volume, c v is always larger than c v . Therefore there will 
be a larger change of entropy for a given ratio of final to initial absolute 
temperature of the system during a constant-pressure than over a 
constant-volume path. This is illustrated in the lesser slope of the 
constant-pressure path of Fig. 6:7. In this figure T z = T 2 , and therefore 
$3 — Si > $2 — Si. The area 1-3-c-a represents the heat flow if constant- 
pressure process 1-3 is carried out reversibly. This area, since c p > c v 
and the constant-pressure and constant-volume processes both take place 
between the same limiting temperatures, must be larger than the area 
under path 1-2. 1 


1 For systems that consist of mixtures of two phases, the specific heat at constant 
pressure is infinite, i.e., heat may be added reversibly at constant pressure without 
change of temperature. For this type of system the constant-pressure process is 
represented by a horizontal line on the TS diagram. 











ENTROPY, A PROPERTY OF THE SYSTEM 


99 


The area under the constant-pressure path on TS coordinates, like that 
under the constant-volume path, has another significance. In Art. 2:8 
it was shown that the heat flow during a reversible constant-pressure 
process was the same as the difference of enthalpy of the system between 
the end states. The area 1-3-c-a of Fig. 6:7 therefore represents this 
change of enthalpy as indicated on the figure. Again, since enthalpy is a 
property, this same area will represent the change of enthalpy for any 
process that connects the same end states. 

Example 6:2 B. A pound of water at 32°F is converted into steam at a temperature 
of 250°F. During the process the pressure is atmospheric. The process consists of 
(a) heating the water to 212°F, ( 6 ) evaporating the water to steam at a constant tem¬ 
perature of 212°F, and (c) heating the steam to the final temperature. During part a 
the specific heat is constant at unity, during part b 970.3 Btu of heat enters the water, 
and during part c the specific heat may be assumed constant at 0.48. Calculate the 
total change of entropy. 

Solution: 


(a) 

(b) 


A 8 0 
AS b 


C p log e 492 

Q b = 970.3 
T 672 


= (1)(0.312) = 0.312 
= 1.445 


(c) A Sc = Cp log e i-H = (0.48) (0.0555) = 0.027 


S 2 - Si = A S a + A S b + A Sc = 0.312 + 1.445 + 0.027 = 1.784 


6:3. The Temperature-Entropy Diagram and the Irreversible Process. 

If, for a simple system, the temperature and the entropy of the system are 
known at a given state, the values of all other properties are fixed and 
may be calculated if the necessary equations of state are available. This 
is true whether the given state has been attained as the result of a reversi¬ 
ble change or as the end state of an irreversible process. Therefore a 
constant-volume irreversible process, for instance, may be plotted, and 
this process will follow exactly the same path as if the process had been 
reversibly performed. Thus path 1-2 of Fig. 6:7 could as well represent 
an equivalent irreversible process for which the succession of states was 
the same as for the reversible process pictured in the figure. However, 
in the case of an irreversible process, the heat flow would no longer be 
represented by the area beneath the path. Instead, the heat flow would 
(in the algebraic sense) be less than this area [see Eq. (6:7) and footnote 
page 95]. The same principle will hold when any property remains 
constant or when the relation between any two independent properties is 
known over the path. 

It has been shown in Chap. 2 that the path of a reversible adiabatic 
process for a given system is quite fixed and definite but that a multitude 
of paths could be traversed if the process were adiabatic but irreversible. 
In Fig. 6:8 are shown a reversible adiabatic process 1-2 and (as 1-3) an 




100 


BASIC ENGINEERING THERMODYNAMICS 


irreversible adiabatic that begins at state 1 and ends at the pressure corre¬ 
sponding to that at point 2. It will be noted that S 3 > S% [see Eq. (6 :/)] 
and that it would not be possible for path 1-3 to slope to the left instead of 

as shown in the figure. The area be¬ 
neath path 1-3 of course has no heat- 
flow significance since this path was 
not traversed during a reversible proc¬ 
ess. We know only that this area must 
exceed the heat flow, and this is of 
course true since the process was speci¬ 
fied to be adiabatic. 

Example 6:3. A system consisting of 1 lb 
of the gas for which equations of state are 
given in Example 1:10 has an initial pressure 
of 60 psia and an initial temperature of 500°F. 
It expands adiabatically but irreversibly to 
final pressure of 10 psia. Because of the 



Fig. 6:8. The reversible and 
irreversible adiabatic process. 


the 


a 


rreversibility the work of the process is 10 per cent less than would have resulted 
from a reversible adiabatic expansion from the initial state to the final pressure, (a) 
Calculate the final temperature for the irreversible process, and compare with the 
temperature at the end of the reversible process. ( b ) What is the change of entropy 
that results from the irreversible process? 

Solution. The notation used will refer to that of Fig. 6 - 8 . Note that 1 Q 2 = 1 Q 3 
= 0 since both processes are adiabatic and that therefore 


Ui — Mj = 


(lW,)r 


and ui 


u 3 = 


1 W, 


J - / 

(a) For reversible adiabatic process 1-2, piVi k = p^v^ 
CT\ (0.37) (960) 


[Eq. (2:3)] 


v x = 


Vi 


60 


[Ex. 2:12A] 

= 5.92 ft-(^ 1 ‘ = 5 Kro) ,/, < 

= (5.92) (3.59) = 21.25 ft 3 


T, = = ■ -°o| 1 7 ' 25 - = 574°R, or 114°F 

Mi = 16 + (0.46)(60)(5.92) = 179.5 Btu;M 2 = 16 + (0.46)(10)(21.25) = 113.8 Btu 
(l W 2 )n 


J 


= Mi — M 2 = 179.5 — 113.8 = 65.7 Btu 


lWd = (0.9)(65.7) = 59.1 Btu; m* = mi - ~ = 179.5 - 59.1 = 120.4 Btu 


J 


J 


m = 120.4 = 16 + (0.46)(10)v 3 or v 3 = = 22.7 ft 3 


(0.46)(10) 


Ti = = — Q Q ( 3 y - 7) = 613°R, or 153°F 


As a result of irreversibility the final temperature increases from 114 to 153°F. 

(5) The change of entropy during a reversible adiabatic process is zero. Therefore 
s 2 = Si and S 3 — Si = s 3 — s 2 . A constant-pressure process connects state 3 with 
state 2 , and the change of entropy is 

















ENTROPY, A PROPERTY OF THE SYSTEM 


101 


S 3 - Si = s, - s 2 = Mc p logo 7 ^ = (1)(0.238) log* = (0.238)(0.0676) 

1 2 574 

= 0.0161 [Eq. (6:11) and Ex. 2:9] 


6:4. The Enthalpy-Entropy (Mollier) Chart and the Steady-flow 
Process. In actual engineering applications of thermodynamic princi¬ 
ples steady-flow conditions are often closely approached. Important 
examples are the heat exchanger and the prime mover that were analyzed 
as steady-flow devices in Chap. 3. As was brought out in that chapter, 
the change of enthalpy during the steady-flow process is of special interest 
and significance, and the enthalpy-entropy diagram (often called the 
Mollier chart in honor of Richard Mollier, who first suggested it) is a 
useful tool in connection with steady-flow analyses. Charts of this type 
are available for most pure substances of engineering significance, specific 
values of the enthalpy (the enthalpy of unit weight) being plotted as 
ordinates against the values of the en¬ 
tropy of unit weight as abscissas. 

A general hs chart is presented in 
Fig. 6:9. A steady-flow heat exchanger 
operates ideally at constant pressure, and 
it would therefore be convenient to have 
lines of constant pressure shown on the 
diagram. For the constant-pressure 
reversible process it has been shown in 
Art. 2:8 that dH may replace dQ. Thus 
ds = dh/T (constant pressure) and 
(dh/ds) p = T. But (dh/ds) p is the slope 
of a line connecting states having equal 



Fig. 6:9. The Mollier chart. 


pressures on the Mollier diagram, and a typical constant-pressure line 
on the enthalpy-entropy diagram will have somewhat the appearance of 
line 1-2 of Fig. 6:9, increasing in slope in proportion to the increase of 
absolute temperature. 1 

Other lines, such as those connecting states of equal temperature, are 
also plotted on the Mollier diagram, and it is easily possible by using them 
to locate any given state of the substance as a point on the diagram. 
Thus, in the case of the steady-flow heat exchanger, points representing 
the state of the fluid at entrance and at exit from that unit may be located 
on the chart, the corresponding enthalpies read from the diagram, and the 
heat flow that has taken place during passage through the heat exchanger 
calculated as the difference of these enthalpies [see Eq. (3:9)]. 


1 In the case of a two-phase mixture heat may be added simultaneously at constant 
pressure and at constant temperature (see footnote, p. 98). For the two-phase 
mixture the line of constant pressure on the Mollier chart would have a constant 
slope and therefore would be a straight line rather than a curve. 






102 


BASIC ENGINEERING THERMODYNAMICS 


The process that takes place in the prime mover of the power plant is 
typically steady-flow and adiabatic. If idealized as a reversible process, 
it would be represented by a vertical line on the Mollier chart (see line 3-4 
of Fig. 6:9). When the state of the fluid at entrance to the prime mover 
is known and some one property other than the entropy (such as the 
exhaust pressure) is established with regard to the condition of the fluid 
at exit from that unit, the state of the fluid at exit may be located. Then 
the maximum work that may be developed per pound of fluid flow may be 
calculated as equal to the decrease of enthalpy which has taken place 
[see Eq. (3:10)]. 

When the steady-flow process as carried out in the prime mover, though 
adiabatic, is not reversible, the effect of that irreversibility is reflected in 
a bending of the path to the right (toward increasing entropies) as has 
already been shown. If the exit state may be established, as at 4a in Fig. 
6:9, then the work per pound of fluid flow may again be calculated by 
applying Eq. (3:10) and is seen to be less than the maximum by the differ¬ 
ence between the enthalpies corresponding to states 4a and 4. Con¬ 
versely, when state 3 and the work per pound of fluid flow are known, 
point 4a may be located. 

The pressure-volume and temperature-entropy diagrams are of maxi¬ 
mum value for the illustration and demonstration of thermodynamic 
principles. When those principles are known and the time has come to 
apply them to the solution of the practical engineering problem, the 
enthalpy-entropy chart is found to be a tool which is excellently adapted 
to that purpose. 

Example 6:4. Assume that the gas of Example 1:10 approaches a steady-flow 
adiabatic prime mover at a pressure of 60 psia and a temperature of 500°F. In the 
prime mover, which, as has been shown in Chap. 3, may be either a turbine or (for 
quasi-steady flow) a reciprocating engine, an expansion takes place to a final pressure 
of 10 psia as in Example 6:3. Neglect differences in velocity and elevation at entrance 
and exit, (a) Assume the expansion to be reversible, and calculate the enthalpies at 
the initial and final states and the work per pound of gas expanded. ( b ) If irreversible 
and the condition of the gas at exit corresponds to state 3 of Example 6:3, calculate 
hz and the work per pound of gas expanded, (c) Compare the work obtained from 
the steady-flow prime mover with the corresponding results of Example 6:3, and 
explain the differences. 

Solution: 


(a) p! = 60 psia; 7\ = 960 o R;v! = 5.92 ft 3 ; hi = 16 + (0.645)(60)(5.92) = 245.2 Btu; 
p 2 = 10 psia; T 2 = 574°R; v 2 = 21.25 ft 3 ; h 2 = 16 + (0.645)(10)(21.25) = 153 Btu. 


(lW 2)steady flow 


J 


= hi - h t = 245.2 - 153 = 92.2 Btu/lb 


(6) p 3 = 10 psia; T 3 = 613°R; v 3 = 22.7 ft 3 ; h 3 = 16 + (0.645)(10)(22.7) = 162.3 Btu. 

(1 IF 3) steady flow 

J 


= hi - h 3 = 245.2 - 162.3 = 82.9 Btu/lb 




103 


ENTROPY, A PROPERTY OF THE SYSTEM 

(c) Comparing the result of part a above with that of part a of Example 6 : 3 , it is 
observed that the work obtained from the expansion of 1 lb of the gas in the reversible 
steady-flow turbine or engine is 92.2 Btu instead of 65.7 Btu as calculated in Example 
6.3. If the device is a turbine, this difference is readily explained (see Chap. 3) as 
the difference between, the flow work done on the open system at entry to the turbine 
and that done by the system at exit, or 

P\Vi P 2 V 2 /144\ 

~j - j~ = [(60)(5.92) - (10)(21.25)] = 26.5 Btu/lb 

Thus 65.7 + 26.5 = 92.2 Btu. 

In the case of the engine, the work that accompanies the reversible adiabatic expan¬ 
sion alone is 65.7 Btu. It is represented by the area 1-2 -d-c under the reversible 
process 1-2 of the figure which accompanies this example. But the engine operates 
continuously, and therefore the fluid system must follow some sort of cycle of events 



that will eventually return it to state 1 . If the pressure is steady at entrance and 
exit, the ideal cycle will be represented by 1-2-6-a of the figure. In this cycle the 
line a -1 represents the induction of the charge into the cylinder while the valve on the 
supply line is open. During this induction, work is performed on the piston equiva¬ 
lent to area a-l-c-0, or P 1 V 1 . Similarly, line 2-6 represents the rejection of the charge 
into the exhaust line. Area 2-6-0-d (a negative area) is the work done by the system 
on the piston during the process; this area is —P2V2. Area 1-2 -d-c has been shown to 
be J ( u\ — U 2 ) = J{ 65.7) ft-lb. No work is done on the piston while it is stationary 
during the process 6 a. Thus the net work of the cycle is 

J(ui — U 2 ) + P 1 V 1 — P 2 V 2 = J (hi — 62 ) ft-lb, or 92.2 Btu/lb 

Note that the engine and the turbine deliver the same maximum amount of work under 
equivalent supply and exhaust conditions. The choice of a prime mover must be based 
on practical factors to be suggested later. 

Now let us turn to the cycle that includes the irreversible expansion which is repre¬ 
sented by the dashed line 1-3 of the figure. The area 1-3-e-c under this curve does not 
represent the work of expansion in this case because the process is irreversible and the 
net work of cycle 1-3-6-a cannot be calculated through the summation of areas as 
before. However, all other processes included in the cycle are reversible, and the 
areas beneath their paths therefore represent the work amounts that accompany them. 
The work performed as the result of irreversible process 1-3 alone has been shown to 
be ui — uz = 179.5 — 120.4 = 59.1 Btu in Example 6:3. Adding the work (in Btu) 









104 


BASIC ENGINEERING THERMODYNAMICS 


equivalent to area a-l-c-0 and subtracting that represented by the area 6-3-a-0, we 
have for the net work of the cycle 


, P iVi 

U\ — Uz -\ - -J- 


P zV 3 

J 


= hi — h 3 (Btu) or 59.1 + 


(144) (60) (5.92) 

778 

(144)(10)(22.7) 

778 


= 82.9 Btu 


as in part b above. 


6:5. The Temperature-Entropy Diagram and the Reversible Cycle. 

In Figure 6:10 is shown a Carnot (reversible) cycle plotted on TS coordi¬ 
nates. The heat supplied per cycle from the source at T s (Qs ) is repre¬ 
sented by the area beneath line 1-2 (area 1-2-6-a) and the heat rejected to 
the refrigerator at T R (Q R ) by the area 3-4 -a-b under line 3-4. By differ¬ 
ence the net work of the cycle is the enclosed area 1-2-3-4. But the effi¬ 
ciency of the reversible Carnot engine that operates on this cycle is the 

proportion which this net work bears to Q s , or area ^^ 4 , anc j a v i sua l 

area l-z-o-ci 

comparison of these areas makes it possible to estimate that efficiency 
without carrying out a detailed calculation. The same principle can be 
used to compare the efficiencies of all reversible cycles easily and effec¬ 
tively whether or not they, like the Carnot, receive and reject heat at 
constant temperature. 


T 




4 

a 


Fig. 6:10. The Carnot cycle on TS 
coordinates. 



Fig. 6:11. The Stirling cycle on TS 
coordinates. 


When regeneration is employed in the course of the cycle, it must be 
taken into account in making a visual estimate of the cycle efficiency. In 
Fig. 6:11 a Stirling cycle is pictured which is equivalent to the Carnot 
cycle of Fig. 6:10 in that it receives heat from the source at the same con¬ 
stant temperature T s and rejects heat to the refrigerator at the same 
constant temperature T R . These two processes are, however, connected 
not by adiabatics as for the Carnot but by constant-volume regenerative 
processes (see Art. 4:10). The crosshatched area 2-3 -c-d represents heat 
rejected to the regenerator and eventually restored during process 4-1 























ENTROPY , A PROPERTY OF THE SYSTEM 


105 


(area 4-1 -b-a ); these heat quantities are neither discharged to the refrig¬ 
erator at T r nor accepted from the source at T s and therefore do not form 
a part of Q R or Q s . Further, it may be shown by applying Eq. (6:10) that 
these areas are equal in size. Qs is drawn from the source during process 
1-2 and is represented by the area 
1 -2 -d-b\ if Ts is the same temperature 
in Figs. 6:10 and 6:11 and if the change 
of entropy S 2 — S\ is also the same in 
both cycles, Q s for the Stirling cycle 
will equal Qs for the Carnot cycle. 

Similarly, if Tr is the same tempera¬ 
ture for both cycles, the area 3-4-a-c 
representing Q R for the Stirling cycle 
can be shown to be equal to area 3-4 -a-b 
of Fig. 6:10 and thus to Q R for the 
Carnot since S s — >S 4 = S 2 — Si for 
both cycles. The cycle efficiencies will 
be equal as has already been demon¬ 
strated. If regeneration had not been brought into play, the efficiency of 
the cycle of Fig. 6:11 would have been less than that of the Carnot as an 
examination of the proportions of the two cycles on the TS diagram will 
show. 

Let us consider and compare the two reversible cycles that are plotted 
together on the temperature-entropy diagram of Fig. 6:12. Both have 
the same maximum temperature of source and the same minimum tem¬ 
perature of refrigerator. Applying the suggestion made earlier in this 
article, it is observed that the efficiency of cycle 1-2-3-4 is greater than 
that of cycle 5-2-3-4 and that the reason it has this advantage lies in the 
higher average temperature at which heat is supplied from the source. 
Another observation is that the efficiency of both cycles would have been 
increased if the temperature T 2 had been higher. A similar gain in 
efficiency would have resulted from a lowering of point 3. It is an 
accepted principle in thermodynamics that the atmosphere forms the 
standard refrigerator of the power cycle and therefore the temperature of 
the atmosphere is the lowest temperature at which heat may be rejected 
to the sink. Assuming T 4 to represent the temperature of the atmos¬ 
phere, if the constant-pressure process 3-4 could have been an isothermal 
at T 4 , Q r would be reduced and the efficiency of both cycles would have 
been improved. For maximum cycle efficiency: 

1 . Heat should be supplied from the source at the highest average tem¬ 
perature that is practicable. 

2 . Heat should be rejected to the refrigerator at the lowest average 
temperature which is consistent with the practical situation. 



g. 6 :12. Comparison of cycles on 
$ coordinates. 






106 


BASIC ENGINEERING THERMODYNAMICS 


6:6. Entropy as a Criterion of Stability. Stability is the characteristic 
of a system that measures its ability to resist changes in its state. It is 
always possible to change the state of a system by allowing it to exchange 
energy with its surroundings. For that reason stability is customarily 
evaluated with reference to a system isolated from its surroundings. For 
a system so situated dQ would be zero, and reference to Eq. (6:7) indicates 
that the entropy of the system could increase or, for the limiting (reversi¬ 
ble) process, remain constant. If the entropy increased, that increase 
must have resulted from an irreversible process and the isolated system 
could not return to its original state. When, as the result of such irre¬ 
versible changes, a state is attained at which the entropy is a maximum 
and no further change in state (except a reversible change) is possible, the 
system is said to be in stable equilibrium. From the First Law comes the 
statement: 

The stored energy of an isolated system will remain constant. Based on 
the Second Law, it may be further stated: 

The entropy of an isolated system may increase but cannot decrease. The 
latter statement is known as the principle of the increase of entropy. When 
combined with the statement based on the First Law, the result may be 
expressed as follows: 

Of all states at which the stored energy of the system is the same , the most 
stable is that corresponding to the greatest entropy. 

For the simple system, internal energy may replace stored energy in the 
preceding statement. 

6:7. Internal Irreversibility. The irreversible processes that were 
the subject of discussion in Art. 6:3 were examples of external irreversi¬ 
bility in that the effects of the factors that cause irreversibility (friction, 
unrestrained expansion, and the flow of heat across a finite temperature 
interval) were in evidence at the boundaries of the system. But it has 
been brought out that even the isolated system may undergo an irreversi¬ 
ble process during which its entropy will increase. This is classified as 
internal irreversibility. Whenever the various parts of a system are not 
at uniform temperature or uniform pressure, an irreversible internal 
change may be expected to result if the necessary disturbance is supplied. 

Consider a continuous system that has been in contact at various sec¬ 
tions of its boundary with external systems at different temperatures. 
As a result the temperature of the system is not uniform. If the external 
systems are removed, the system becomes an isolated system. Its state 
cannot be plotted as a single point on the temperature-entropy diagram, 
but it may be divided into many small systems each of which has a uni¬ 
form temperature and each of which is therefore in an equilibrium state. 
The states of these infinitesimal subdivisions of the entire system will not 
persist, however, for the effects of molecular activity will be such as 


ENTROPY , A PROPERTY OF THE SYSTEM 


107 


eventually to equalize their temperatures. The process is not reversible 
since the external systems must again be called on to restore the original 
temperature differentials and the entropy of the system must have 
increased while it was isolated. The change of entropy may be calculated 
if sufficient information is available. 

Example 6:7 A. A system consists of a block of copper (specific heat 0.0915), 
weighing 200 lb and having a temperature of 200°F, and a bucket containing 100 lb 
of water (specific heat unity) at a temperature of 60°F. What change of entropy will 
result when the copper is immersed in the water? Neglect radiation and the thermal 
capacity of the bucket. 

Solution. The internal energy of the system does not change. The energy 
exchanged between the copper and water is 

(200) (0.0915) (200 - t) = (100)(1)(« - 60) or t = 81.6°F 

in which t is the final temperature of both the copper and the water. The change 
of entropy of the copper is 

(200)(0.0915) loge (541.6/660) = (200)(0.0915)(-0.198) = -3.62 
The change of entropy of the water is 

(100)(1) loge (541.6/520) = (100)(0.0412) = 4.12 
The change of entropy of the entire system is 

-3.62 + 4.12 = 0.5 

Example 6:7 B. A system consists of one pound of a gas with equations of state 
as in Example 1:10 at a pressure of 60 psia and a temperature of 200°F and a second 
pound of the same substance at the same pressure but at a temperature of 60°F. The 
system* is confined within a rigid nonconducting container, the two parts being sepa¬ 
rated by a thin but rigid partition through which thermal energy passes until the 
temperatures of the two parts of the system are equalized. ( a ) At the end of this 
process, what has been the change of entropy of the system? (6) If, after the tem¬ 
perature has been equalized, the partition is punctured, will any further change of 
entropy result? 

Solution: 

(a) The final temperature of both parts of the system will be (200 + 60)/2 = 130°F. 
The specific heat at constant volume for this gas is 0.17 (see Example 2:7), and the 
change of entropy for the system is 

(1)(0.17) log, + (1)(0.17) log, Ifl = (0.17)(-0.0675 + 0.1267) = 0.0101 

( b) At the end of the first process and before the partition is punctured, the pressures 
of the two parts of the system are unequal. The pressure of the part originally at 
higher temperature is (590)(60)/660 = 53.7 psia (its volume has remained constant). 
Similarly, the pressure of the second part is (590) (60)/520 = 68.1 psia. After the 
partition was punctured, these pressures were equalized at 60 psia while the tempera¬ 
ture remained at 130°F. At this common pressure and temperature the system would 
have been in stable equilibrium even if the two parts had still been separated by the 
partition. We may therefore calculate the change of entropy during this second 
process as if the partition had not been punctured but had moved along the cylinder 



108 


BASIC ENGINEERING THERMODYNAMICS 


as a piston might move; as a result of this movement the part at higher pressure 
would expand isothermally, and that at lower pressure would be compressed isother- 
mally until their pressures were equalized. Equation (1:6) indicates that for these 
isothermal processes the product pv would not change and that therefore, by Eq. (1:7), 
there would be no change of internal energy of either part. Then, by Eq. (2:5), 
dQ = dW/J = P dV/J. But pv is constant, and P = W^p\V\/v = 144CT i/v. Then 


As = 



f2 Pdv 
l JT 


144 CT 
778 T 


[ V2 — = 0.185C log, - = 0.185C log 
Jv 1 V Vi 


Pi 

P* 


Applying this equation to each of the two parts of the system, the total change of 
entropy is 


(0.185)(0.37) [log, (60/53.7) + log e (60/68.1)] = 0.0685(-0.1115 + 0.1269) = 

+0.00105 


The total change of entropy as the two parts of the system are brought into tempera¬ 
ture and pressure equilibrium is 0.0101 + 0.00105 = 0.01115. 

6:8. The Availability and Unavailability of Heat. It has been shown 
that not all of the heat taken from a source may be continuously changed 
to work, that the device which will change the maximum proportion of 
heat so received into work is the reversible engine of which the Carnot 
engine is an outstanding example, and that the proportion changed into 
work by the Carnot engine will depend on the temperatures of the source 
and of the refrigerator. It has also been brought out that the tempera¬ 
ture of the refrigerator is, ultimately, that of the atmosphere and may 
therefore be considered a constant temperature. If the source is capable 
of supplying heat at constant temperature, the maximum portion of the 
heat it supplies that can be converted into work, as the result of a cyclic 
process with heat rejection to the atmosphere, may be readily calculated 
as the ratio of area 1-2-3-4 of Fig. 6:10 to area 1-2 -b-a of the same figure. 
In thermodynamics this is called the available portion of the heat supplied 
by the source. Similarly, the ratio of area 4-3 -b-a to area 1-2-6-a of the 
figure is that portion of the heat supplied by the source which not even the 
ideal Carnot engine could transform into work and is thus the minimum 
portion that must be rejected to the atmosphere (the refrigerator); this is 
termed the unavailable portion. 

It has been inferred above that the source was infinite in extent and 
thus capable of supplying heat at constant temperature over an indefinite 
period of time. But it may be possible for even a finite system to act as a 
source in supplying heat at temperatures above that of the atmosphere, 
although, being finite in extent, such systems often change in temperature 
as they receive or emit heat. Suppose path 1-2 of Fig. 6:13 to represent a 
reversible process during which the system receives heat as its tempera¬ 
ture changes. A small increment of heat produces only an infinitesimal 
temperature change and, as shown in the figure, can supply a small Carnot 
engine that rejects heat to the atmosphere at T 0 . When the effects of all 



ENTROPY, A PROPERTY OF THE SYSTEM 


109 


these little Carnot cycles are combined, it is evident that the total heat 

they have constructively received is T dS and is represented by the 

area 1-2 -d-c under the path. This heat which has been received by the 
system is divided into two parts. 

The upper area 1-2-5-a is the sum of 
the enclosed areas of the Carnot cycles 
and represents the maximum work 
that could be produced from this heat 
as the result of a cyclic process; it is 
that part which is available. The 
lower area a-b-d-c, or T 0 (S 2 — S i), is 
the sum of the heat amounts rejected 
to the atmosphere by the Carnot 
cycles and represents the minimum 
amount of heat that would be rejected 
to the atmosphere as work is developed 
due to a cyclic process; it is that part 
which is unavailable. Thus the available part of the total heat flow is 

^ T dS — T 0 (S 2 — Si), and the unavailable portion is T 0 (S 2 — Si). 



Fig. 6:13. 
source. 


Availability of heat-finite 


Example 6:8A. One thousand Btu of heat leaves the hot gases in a boiler firebox 
at a constant temperature of 2000°F and enters the water in the boiler, evaporating it 
into steam at a constant temperature of 400°F. The temperature of the atmosphere 
is 80°F. (a) What is the change of entropy of the hot gases? ( b ) Of the water? 

(c) What part of the total heat that leaves the hot gases is available? What part is 
unavailable? ( d ) Divide the heat that enters the water into available and unavail¬ 
able portions, (e) Considering the hot gases and the water to constitute a single 
system, calculate the change of entropy of this system as the result of the transfer of 
thermal energy between its parts, and compare the product T 0 (S 2 — Si) with the 
change of unavailable energy as calculated in (c) and ( d ) above. 

Solution: 


(a) AS (gases) = - = -0.406 

(b) AS (water) = -^inr = 1.162 

(c) Available portion (gases) = 1000 f ^ ~ 1000 ^ ^^460^^ ) _ 7go Btu 

Unavailable portion (gases) = 1000 — 780 = 220 Btu 

(d) Available portion (water) = 1000 = ^72 Btu 

Unavailable portion (water) = 1000 — 372 = 628 Btu 

(e) AS = -0.406 + 1.162 = +0.756; T 0 (S 2 - Si) = (540)(0.756) = 408 Btu 

This agrees with the change of unavailable energy from (c) and ( d ) since 628 — 220 
= 408 Btu. 

Example 6:8 B. One pound of the gas of Example 1:10 changes at constant pressure 
of 60 psia from t\ = 200°F to t 2 = 600°F. The temperature of the atmosphere is 
60°F. (a) Assuming the process to be reversible, what is the heat flow? (6) What 

part of this heat is available? Unavailable? 













110 


BASIC ENGINEERING THERMODYNAMICS 


Solution: 

c P = 0.238 [Ex. 2:9] 

(а) ,Q 2 = Mc p {t 2 - h) = (1) (0.238) (600 - 200) = 95.2 Btu 

T 10fiO 

(б) s 2 - s, = Cplog,^ = 0.238 log,^ = (0.238) (0.474) = 0.113 

Unavailable portion = To(s 2 — Si) = (520) (0.113) = 58.7 Btu 
Available portion = 95.2 — 58.7 = 36.5 Btu 

6:9. The psi function, also called the Helmholtz free-energy function, 
is a composite and extensive property of the system that is defined by the 
expression 

T = E - TS (6:12) 


For the simple system ., E = U, and we may write 


T = U - TS 

or, for unit weight, 

\p = u — Ts 


(6:13) 

(6:14) 


Maximum work is always associated with the reversible process. 
When the simple system undergoes an isothermal and reversible process, 
the decrease of the psi function is 


- *2 = Hr - U 2 - T(S X - S 2 ) = -A U + T AS (6:15) 
But we may also write for this process 
(iW I) n 


J 


= -AU + (iQ 2 )max = -A U + T AS = *1 - *2 (6:16) 


It is observed that the decrease of T will measure the maximum possible 
work that can be performed by the system when the process takes place at 
constant temperature. 

If the system exchanges heat only with the atmosphere, the maximum 
value that 1 Q 2 may have is To(S 2 — S 1 ). Returning to Eq. (2:3), 

(l W 2 )n 


J 


= Ui- U 2 + GQ 2 W = Hi - Ut + To(S 2 - S{) 

= (U 1 - T 0 S 1 ) - {U 2 - T 0 S 2 ) 
= (6:17) 


We may therefore conclude that the decrease in the value of 'Y for the system 
between any two states in each of which it is in temperature equilibrium with 
the atmosphere will measure the maximum work that the system can produce 
during any process which connects those states provided that heat can be 
exchanged only with the atmosphere. That this statement is valid may be 
shown by a return directly to the Second Law. For if a process that con¬ 
nects end states as described above and during which the system would 
produce an amount of work greater than Ti — T 2 could be found, then a 





ENTROPY, A PROPERTY OF THE SYSTEM 


111 


cycle could be formed by returning the system to the original state over a 
reversible isothermal path and net work could be developed at the expense 
only of energy taken from a single reservoir at uniform temperature. 

6:10. The zeta property, or Gibbs free-energy function, is defined as 

Z = E+~~TS (6:18) 

When energy storage due to motion, gravity, capillarity, magnetism, and 
electricity is negligible, this may be rewritten as 

PV 

Z=U + -j--TS = H-TS (6:19) 

or, for unit weight, 

£ = u + ^j — Ts = h— Ts (6:20) 

The zeta function has much the same relation to the open system and 
the steady-flow process as has the psi function to the closed system and 
the nonflow process. Work that is done on the surroundings, such as the 
net flow work performed by the open system at points of entry and exit, 
was differentiated (see Chap. 3) from the useful work associated with the 
flow process. The amount of the corresponding deduction from the 
gross work is, for steady flow, P 2 v 2 — P\V\ per pound of fluid passing the 
boundaries. The decrease of £ is 

— ^2 — ci — e 2 H— -j - j - (T \S\ — T 2 s 2 ) 

, , P2V2—P1V 1 fct 

= Yi — V2 -j- (6:21) 

The preceding article has made it clear that 2 measures the maxi¬ 

mum work that can be produced by the closed system of unit weight dur¬ 
ing a process which proceeds at constant temperature. When an open 
system and a steady-flow process (also at constant temperature) are 
involved, the maximum useful work that may develop can be found by 
deducting the net flow work from this gross amount. An examination of 
Eq. (6:21) will show that the right side of the equation indicates just such 
a deduction. We may therefore state that the decrease in the value of f will 
measure the maximum useful work per pound of fluid flow that can result 
from a steady-flow constant-temperature process which connects the states at 
which f is measured. 

For an open system to be in equilibrium with the atmosphere, it is 
required that at points of entry and exit both the temperature and the 
pressure shall correspond with atmospheric temperature and pressure. 







112 


BASIC ENGINEERING THERMODYNAMICS 


Substituting T 0 for 7\ and T 2 and replacing Pi and P 2 by P 0 , Eq. (6:21) 
becomes 

fi — £2 = £i — ^2 H--"j-- — ? 7 o(si ~ $ 2 ) (6:22) 

This equation states the maximum useful work that could be developed 
as the result of a constant-temperature process connecting states in each 
of which the system is in temperature and pressure equilibrium with the 
atmosphere. But, returning to the Second Law as in Art. 6:9, it becomes 
clear that this maximum applies also to any process which connects those 
states if there has been no heat exchange except with the atmosphere. 
Thus the decrease in the value of Z between any two states in each of which it is 
in temperature and pressure equilibrium with the atmosphere will measure 
the maximum useful work that the system may produce during any process 
which connects those states provided that heat may be exchanged only with the 
atmosphere. 

Although the concept of the open system has been employed in its 
development, the foregoing principle is valid for the closed system also. 
For the work done on the atmosphere must be deducted from the total 
work to obtain the useful work; this deduction is P 0 (V 2 — Vf), and com¬ 
parison will show that it is equivalent to the addition of the second term 
on the right side of Eq. (6:22). 

The zeta function finds its most useful applications when chemical 
changes take place in the course of the contemplated process and the sys¬ 
tem consequently cannot be treated as composed of a pure substance. 
This explains why the more general term e replaces u in Eqs. (6:21) and 
(6:22). An example of its use is in connection with an analysis of the 
performance of an internal-combustion engine. The maximum work 
that can be delivered to the piston is equal to the decrease of Z between 
the original state of the mixture of fuel and air at the pressure and tem¬ 
perature of the atmosphere and the final state corresponding to pressure 
and temperature equilibrium of the products of combustion with the 
atmosphere. The zeta function finds numerous applications in the 
thermodynamics of chemistry. 

When the system is the same pure substance at the beginning and end 
of the process, e\ and e 2 in Eq. (6:22) may be replaced by Ui and u 2 . In 
general, for that case, the internal energy, the volume, and the entropy 
are functions of p 0 and TV Then U\ = u 2 = u 0 , Vi = v 2 = v 0 , Si = s 2 = 
So, and £1 = £2 = To-* This is called the dead state, since no useful work 

* It is assumed here that the storage of energy due to motion, gravity, capillarity, 
magnetism, and electricity is not necessarily absent but is negligible. An exception 
to the general case is found in two-phase mixtures for which the pressure and the 
temperature are not independent properties. For such mixtures the initial and final 
values of the internal energy, volume, and entropy need not be equal. But even in 



ENTROPY, A PROPERTY OF THE SYSTEM 


113 


may be obtained from the system. For the dead state 


— e 0 + 


P 0^0 
J 


T oSo 


Both the psi and the zeta properties may be compared with enthalpy in 
the sense that the change of a property measures the amount of a transi¬ 
tory quantity during a specified process. For it will be remembered that 
the change of enthalpy measured the heat flow during a constant-pressure 
nonflow reversible process and therefore the maximum heat flow, just as 
the decrease of psi and of zeta measures the maximum work that can 
accompany a constant-temperature process during a nonflow and a flow 
process, respectively. They differ from enthalpy, however, in that the 
concept of availability is introduced in both the decrease of psi and the 
decrease of zeta. 

In addition to the uses specifically suggested above, the zeta function 
has great value as a basis for the evaluation of the maximum work that a 
system at any state may, in combination with an atmosphere at pressure 
Po and temperature To, deliver for useful purposes to external systems. 
This application of the zeta function will be demonstrated in a later 
chapter. 

Problems 


1. Which of the following statements are true? (a) (j) ^ = 0 for a reversible 

cycle. (6) (^) ^ = 0 for any cycle, (c) (j) dS = 0 for a reversible cycle, (d) (j) dS 
— 0 for any cycle. 

2. In Art. 1:8, two simple cycles were described, both following the same state path. 
For the first, fdW = 0; for the second fdW < 0. Determine, for each cycle, 

whether (j) ^ is equal to, greater than, or less than zero; also, whether (j) dS is equal 


to, greater than, or less than zero. 

3. During a reversible process that takes place at a constant temperature of 300°F, 
300 Btu of heat flows into a fluid system, (a) What is fdQ/T for this process? (6) 
What is JdS ? The same state path may be followed if the system is stirred while 
adding only 200 Btu of heat. For this second process, (c) what is / dQ/T? ( d ) What 
is fdS? 

4. Work Example 6:1C, but substitute (a) gas W; (b) gas X; (c) gas Z. Are the 
changes of entropy the same for all four gases? 

5. One pound of (a) gas W, ( b ) gas X, ( c ) gas Y, (d) gas Z changes from a state in 
which its pressure is 100 psia and its temperature is 400°F to a second state in which 
these properties are 80 psia and 300°F. Choosing any convenient reversible path, 
calculate the change of entropy. Check your answers by making a similar computa¬ 
tion over some other reversible path that connects the same end states. 

6 . In tables that present the properties of water and steam, the entropy is shown as 
zero for liquid water at 32°F and a pressure of 0.088 psia. The specific entropy of 


this case fi = £2 = To since it will be found that the first two terms of Eq. (6:22) will 
be balanced by the last term and the right side of that equation will still be zero. 




114 


BASIC ENGINEERING THERMODYNAMICS 


steam at the same temperature and pressure is tabulated as 2.1877. What qualifica¬ 
tion applies to the tabulation of this value as the entropy? Assuming that it is 
possible to find a reversible isothermal process which connects the two states, what 
heat flow will take place in the course of that process? Could the same change in 
state be accomplished by a process that is accompanied by a different heat flow than 
you have calculated above? Could the same change of state be accompanied by a 
different change in entropy than is shown in the tables? 

7. Liquid water at 32°F and 0.088 psia (see Prob. 6) may be changed reversibly to 
liquid water at 212°F, 14.7 psia, during a process for which the heat-flow rate is very 
nearly constant at 1 Btu/lb for each °F increase in temperature. What is the tabu¬ 
lated value of the entropy for liquid water at 212°F, 14.7 psia? 

8. When plotted on an absolute temperature vs. entropy diagram, the area under 
the state path (between the path and the S axis) of a certain reversible process, during 
which there is a continuous decrease in system entropy, is 3.1 in. 2 The scale of tem¬ 
peratures on the diagram is 1 in. = 200°R, and the scale of entropy is 1 in. = 0.4 
rank. What heat flow accompanies the process? If the process took place at an 
average temperature of 240°F, what is the difference between the initial and final 
entropies of the system? 

9. A reversible process, during which there is a continuous increase in system 

entropy amounting, in total, to 0.3 rank, is plotted on a temperature-entropy diagram 
for which the origin is located at 0°F. The area between the state path and the S 
axis is 2.4 in. 2 The scale of ordinates is 1 in. = 100°F and, of abscissas, 1 in. = 0.5 
rank. What is the heat flow during the process? 1 

10. In Example 2:10, what is the change of entropy of the system? 

11. During an isothermal process, the entropy of a system composed of an ideal gas 
increases by 0.3 rank. The process takes place at 200°F. What maximum work may 
accompany the process? If the system had not been composed of an ideal gas, would 
the solution of this problem have been possible? Could the maximum heat flow have 
been calculated in both cases? 

12. A certain adiabatic process takes place at constant internal energy. Based 
on Eq. (6:7) and (6:9), does the volume of the system increase or decrease? 

13. At a certain state, a gas system has a pressure of 100 psia, a volume of 1 ft 3 , and 
a temperature of 400°F. Find the volume and temperature corresponding to a state 
having the same entropy but a pressure of 40 psia, if the system is composed of (a) 
gas W ; ( b ) gas X ; (c) gas Y ; ( d ) gas Z. 

14. Find the change of entropy for a 1-lb gas system as the result of a constant- 
volume process during which its pressure is halved if the system is composed of (a) 
gas W ; (6) gas X; (c) gas Y ; ( d ) gas Z. 

15. A system is composed of a substance having a constant specific heat at constant 
volume of 0.4 Btu/(°F)(lb). Its entropy increases by 0.2 rank, its temperature from 
50 to 200°F, during a constant-volume process. What is its weight? 

16. Find the change of entropy of a 3-lb gas system during a constant-pressure 
process which doubles its volume if the system is composed of (a) gas IF; (6) gas X; 
(c) gas F; ( d ) gas Z. 

17. The state path followed by a system during a certain constant-volume process 
is plotted on an absolute temperature-entropy diagram, and the area between the 
path and the S axis is found to be equivalent to 80 Btu. The heat flow during the 
process is 70 Btu. Which of the following statements are justified? (a) The change 
of internal energy of the system is 80 Btu. ( b ) The change of internal energy is 70 
Btu. (c) During the process the internal energy increased. ( d ) The heat flow was 
negative. ( e ) The statement of the problem is a contradiction of thermodynamic 
principles. 


ENTROPY, A PROPERTY OF THE SYSTEM 


115 


18. The state path followed during an irreversible constant-pressure process is 
plotted on an absolute temperature-entropy diagram. Which of the following state¬ 
ments are true? (a) The area between the state path and the S axis is proportional 
to the heat flow. (6) The area is proportional to the change of enthalpy of the system, 
(c) If the entropy decreased during the process, the enthalpy must necessarily have 
decreased. 

19. Work Example 6:3, but use a 1-lb system composed of (a) gas W ; (6) gas X ; 
(c) gas Z. 

20. Show that, for an ideal gas, a line representing a constant-temperature change of 
state would be a horizontal line on the Mollier diagram. 

21. The pressure and temperature of a gas is 100 psia and 400°F as it enters an 
adiabatic steady-flow expansion to a pressure of 20 psia, during which its entropy 
increases by 0.03 rank/lb. Assuming negligible differences in velocity and elevation 
at entrance and exit, calculate the external work performed per pound of gas flow if 
the gas is (a) gas W; (6) gas X; (c) gas Y. 

22. A gas approaches a steady-flow adiabatic prime mover at a pressure of 75 psia 
and a temperature of 500°F. The gas leaves the prime mover at a pressure of 15 psia. 
There are negligible differences between the velocities and the elevations at entrance 
and exit. The work performed is 10 per cent less than the maximum possible in 
adiabatic steady-flow expansion between the initial state and the final pressure. 
What is the final temperature if the gas is (a) gas W; (b ) gas X; (c) gas Z? What are 
the differences between the specific entropy at entrance and at exit for each of the 
gases ? 

23. A reversible closed-system cycle, when plotted on temperature-entropy coordi¬ 
nates, encloses an area of 2.5 in. 2 The scale of temperatures is 1 in. = 250°F; the 
scale of entropies is 1 in. = 0.5 rank. What is the net heat flow for the cycle? The 
net work? 

24. A reversible cycle of an ideal-gas system consists of the following four processes 
in the order given: (1) an adiabatic compression from 100 to 700°F; (2) a constant- 
volume process ending at 2000°F; (3) an adiabatic expansion to 730°F; (4) a constant- 
volume process to the starting point. Plot this cycle approximately to scale on a Ts 
chart, and estimate the cycle efficiency by a visual comparison of areas on the diagram. 

25. The same as Prob. 24, except that the four reversible processes which, in the 
order given, constitute the cycle are (1) an adiabatic compression from 100 to 975°F; 
(2) a constant-pressure expansion to 2000°F; (3) an adiabatic expansion to 730°F; 
(4) a constant-volume process which returns the system to its starting point. Based 
on a visual inspection of the two diagrams, how does the efficiency of this cycle com¬ 
pare with the efficiency of the cycle of Prob. 24? 

26. On the diagrams drawn in the course of the solution of Probs. 24 and 25 super¬ 
impose a Carnot cycle which is supplied with heat from a source at a temperature equal 
to the highest temperature at which heat is supplied in either of the two cycles and 
which rejects heat to a sink at a temperature equal to their lowest temperature. Esti¬ 
mate visually the efficiency of this Carnot cycle, and check by calculation. How 
does its efficiency compare with the efficiencies of the cycles of Probs. 24 and 25? 

27. Prove, based on the principle of the increase of entropy and Eq. (6:8), that the 
constant-internal-energy process described in Art. 2:11 and illustrated in Fig. 2:7 
can take place in one direction only. 

28. A system consists of a 50-lb iron ball (specific heat = 0.101) at a temperature of 
200°F and a 100-Z6 bath of water (specific heat = 1) at 60°F. The ball is immersed in 
the water, and the two reach a common temperature. The exchange of energy with 
external systems is negligible. Prove, based on the principle of the increase of 
entropy, that the process is irreversible. 


116 


BASIC ENGINEERING THERMODYNAMICS 


29. A pound of liquid water (specific heat = 1) at 70°F is heated at constant atmo¬ 
spheric pressure (14.7 psia) to 212°F and is then evaporated into steam as 970.3 Btu of 
heat is added at constant temperature. The temperature of the atmosphere is 70°F. 
What fraction of the heat expended in increasing the temperature of the water is 
available? What fraction of the heat supplied in evaporating the water into steam 
is available? What is the total change of available energy of the system? Of 
unavailable energy? 

30. In Prob. 28, what is the change of available energy of the system if the tem¬ 
perature of the atmosphere is 50°F? What is the change of unavailable energy? 

Symbols 

c p specific heat at constant pressure 
c v specific heat at constant volume 
C a constant 

e stored energy of unit mass, in general 
E stored energy of a system in general 
h enthalpy of unit mass 
H enthalpy of a system 
J proportionality factor 
k a constant ratio 
M mass 
p pressure, psi 
P pressure, psf 
Q heat flow 
s entropy of unit mass 
S entropy of a system 
t scalar temperature 
T absolute temperature 
u internal energy of unit mass 
U internal energy of a system 
v volume of unit mass 
V volume of a system 
W work 

Greek Letters 

£ zeta, or Gibbs free-energy, property of a unit system 
Z zeta, or Gibbs free-energy, property of a system 
\p psi, or Helmholtz free-energy, property of a unit system 
psi, or Helmholtz free-energy, property of a system 

Subscripts 

p constant pressure 
R refrigerator 
S source 

v constant volume 


CHAPTER 7 


THE PURE SUBSTANCE 

7:1. Phases of the Pure Substance. All pure substances follow much 
the same behavior pattern in passing through their various phases. At 
relatively low temperatures they are characteristically solids, and as the 
temperature rises they melt, or fuse, to become liquids. Finally, still 
farther up the scale of temperature, they vaporize into gases or vapors. 
The temperatures at which these transitions take place are not, for a 
given substance, fixed and definite but are functions of the pressure. The 
melting and boiling temperatures of different substances differ so widely 
that we are accustomed to speaking of them as gases, as liquids, or as 
solids according to their phase under the conditions at which we usually 
observe them. Thus iron is thought of as a solid, mercury as a liquid, 
and oxygen as a gas because at usual atmospheric temperatures and pres¬ 
sures they are observed in these phases. Water, on the other hand, is a 
substance which we observe in all three of these phases. Because of this 
and since, as perhaps the most important of thermodynamic pure sub¬ 
stances, the amount of research that has been centered on the determina¬ 
tion of its properties over a wide pressure-temperature range is undoubt¬ 
edly greater than for any other substance, water is a convenient substance 
on which to base our study of the behavior of pure substances in general. 

Let us begin with water in its solid phase, or as ice, and let us assume 
that we are dealing with 1 lb at a temperature of 0°F and at standard 
atmospheric pressure. We may raise the temperature of this ice by add¬ 
ing heat to it, and if we can devise a method of measuring this heat, it will 
be found that approximately 16 Btu will have been supplied as the tem¬ 
perature increases to 32°F. Since the heat has been supplied at constant 
pressure, its amount measures the change of enthalpy of this unit weight 
of ice. The specific heat at constant pressure is therefore approximately 
0.5 Btu/(lb)(°F). 

It will be remembered that this temperature of 32 deg was one of the 
two anchor points on the Fahrenheit scale of temperature and was, 
specifically, the melting point of ice under 1 standard atmosphere. We 
therefore expect to find a change in phase taking place that will eventually 
convert all of the ice into the liquid form which we call water. This con¬ 
version takes place at the constant temperature of 32°F and requires that 
144 Btu of heat be supplied to change the pound of ice completely to the 

117 


118 


BASIC ENGINEERING THERMODYNAMICS 


form of a liquid. Heat supplied under conditions such as this is often 
referred to as the latent heat of fusion, the word latent being used to indi¬ 
cate that the effect of the addition of this heat is not visible in terms of a 
change of temperature. The use of latent heat of fusion may be criticized 
from the thermodynamic standpoint since melting could have been 
accomplished by supplying energy in the form of work. It will be noted 
that the pressure was, by assumption, constant during the melting process 
and that this 144 Btu therefore represented the increase of enthalpy dur¬ 
ing fusion; a better thermodynamic designation would therefore be the 
change of enthalpy during fusion or simply the enthalpy of fusion. 

By increasing the pressure acting on the surface of the ice and water, 
we could have decreased the temperature at which fusion took place. 
The rate of decrease of melting temperature with increase of pressure is 
quite small, but, by increasing the pressure to some 2000 atm (about 
30,000 psia), the temperature of fusion can be lowered to a minimum value 
of about — 8°F. Above this pressure the ice assumes a different crystal¬ 
line structure, and the melting point rises. This new form of ice may be 
classified as an entirely separate and distinct phase. Bridgman 1 has 
described a total of seven different solid phases of water and has demon¬ 
strated that fusion temperatures above 180°F are possible. The ice 
phase with which we are familiar is designated as ice I, the others, simi¬ 
larly, as ice II to ice VII. Of these, ice II cannot be made to change 
directly into the liquid phase. 

Returning to the lower anchor point of the Fahrenheit scale and assum¬ 
ing that the pressure is again constant at 1 standard atmosphere as we 
supply heat to the pound of water that has resulted from the melting of 
the ice, we observe that the temperature will again rise as we add this heat 
until the upper anchor point of 212 deg is reached. During the interval 
of 180 deg we shall have supplied some 180 Btu, and the mean specific 
heat is therefore (very nearly) unity. As we add further heat quantities, 
no further change in temperature is immediately observed; instead the 
liquid boils and progressively changes its form to a gas. To accomplish 
the complete conversion at this temperature and pressure into 1 lb of the 
gas ( steam, or water vapor ) requires the addition of some 970.3 Btu. This 
is the latent heat of evaporation, or, more properly, the enthalpy of vaporiza¬ 
tion at this pressure and temperature. After the liquid has been com¬ 
pletely converted into the gaseous phase, a further supply of heat will 
cause the temperature to resume its upward march and the fluid becomes 
a superheated vapor, or gas. A distinction is sometimes made between 
these terms based on the number of degrees of superheat, but no definite 
and distinct boundary exists between the superheated vapor and the gas, 
and the two terms are often used synonymously. 

1 J. Franklin Inst., 177, 315(1914); J. Chem. Physics, 5, 964(1937). 


THE PURE SUBSTANCE 


119 


It has been noted that the fusion temperature of ice is a function of the 
pressure applied on its surface. The same is true of the boiling tempera¬ 
ture of water. If the pressure on its surface had been 100 psia, for 
instance, instead of 1 standard atmosphere, the temperature of the water 
would have risen to about 327.8°F before it began to boil. To raise its 
temperature to this level from 32°F would have required that some 298.4 
Btu of heat be supplied to the pound of water. Evaporation from and at 
this temperature (and under a constant pressure of 100 psia) would have 
required 888.8 Btu, the enthalpy of vaporization under these new condi¬ 
tions; it will be noted that, though the investment of heat in the liquid 
has been greater than for evaporation at 212°F, the enthalpy of vaporiza¬ 
tion has decreased. Conversely, if the pressure were 5 psia, the boiling 
temperature would be 162.2°F, the enthalpy of the liquid 130.1 Btu, and 
the enthalpy of vaporization 1001.0 Btu/lb. 

The observed trend toward the decrease of the enthalpy of vaporization 
with increased temperature and pressure is consistent and is characteristic 
of all pure substances. It reaches its logical end point when the enthalpy 
of vaporization becomes zero. For water this comes at a temperature of 
about 705°F and a corresponding pressure of about 3206 psia. This is the 
so-called critical point . At each temperature between the melting point 
and the critical point a combination of pressure and temperature may be 
found at which the liquid and the vapor phases may be in pressure and 
temperature equilibrium and mixtures of the two phases may therefore 
exist in such equilibrium. Above the critical temperature this is not 
possible. If mixtures of the two phases exist above the critical tempera¬ 
ture, they cannot be in equilibrium and will persist only so long as it takes 
for energy transfers to change the liquid into the vapor, the vapor into the 
liquid, or to bring the entire mixture below the critical temperature. The 
statement is often made that the critical temperature is the highest tem¬ 
perature at which the substance may exist in the liquid phase. This is 
subject to some criticism, for, at some temperatures and pressures in 
excess of the critical values of these properties, the characteristics of the 
substance are such that there is some doubt as to whether it should be 
classified as liquid or as gas. In this “twilight zone” it is perhaps better 
to avoid the issue by applying the term fluid to the condition of the sub¬ 
stance. It may be safely said that the critical temperature is the highest 
temperature at which mixtures of the liquid and vapor phases may be in 
equilibrium. 

A vapor that is in pressure-temperature equilibrium with its liquid is 
called a saturated vapor , the liquid a saturated liquid. Mixtures of the two 
will have extensive properties that will be dependent on the proportions 
of the vapor and the liquid in the mixture. For such mixtures the pres¬ 
sure and temperature are not independent properties, and the state of the 


120 


BASIC ENGINEERING THERMODYNAMICS 


mixture cannot be determined without knowledge concerning at least one 
additional property. This property is often the quality. Quality is 
defined as the proportion of the vapor in the mixture by weight or the 
weight of vapor contained in 1 lb of the mixture and is conventionally 
designated by the symbol x. Any extensive property of such mixtures is 
the sum of the corresponding properties of its parts so that the total value 
of the property for the mixture of unit total weight would be x times the 
value of the property for the saturated vapor at the specified temperature 
(or pressure, for the pressure establishes the temperature, and vice versa) 
plus (1 — x) times the specific value of the same property for the saturated 
liquid at that temperature. 

It has been observed that for each saturation temperature up to the 
critical temperature there is a corresponding saturation pressure. As we 
move downward along the scale of saturation temperatures, we find that 
there is a corresponding decrease in saturation pressures. At 32°F we 
find that the saturation pressure for water is about 0.089 psia. But the 
fusion temperature for ice, it will be remembered, increases with lowered 
pressure and was 32°F at 14.7 psia, 1 standard atmosphere. At a pressure 
of 0.089 psia the fusion temperature is 32.02°F. The intersection of the 
line of fusion temperatures with the line of saturation temperatures is 
called a triple point because at the corresponding pressure of 0.089 psia 
and temperature of 32.02°F the solid, liquid, and vapor phases of water 
may coexist in equilibrium. 

Below the triple-point temperature the liquid cannot be saturated, i.e. f , 
cannot be in pressure-temperature equilibrium with the vapor. By 
increasing the pressure, equilibrium may be maintained at lower tempera¬ 
tures between the solid and the liquid, as we have already observed. 
Similarly, by decreasing the pressure below 0.089 psia it is possible to 
maintain equilibrium between the solid and the vapor phases so that 
mixtures of these two phases at a common temperature and pressure are 
possible. If heat is supplied at constant pressure and temperature to 
equilibrium mixtures of the solid and vapor at temperatures below the 
triple-point temperature, the solid will change progressively directly into 
the form of a vapor without passing through any intermediate liquid 
phase. The heat required to accomplish the change of unit weight of the 
solid into the vapor under these conditions is called the heat of sublimation 
or, better, the enthalpy of sublimation. When wet clothing is exposed to 
air well below freezing temperatures, it will be observed to freeze and then 
to dry directly from this condition without again becoming “wet.” 
This occurs when the air is “dry,” i.e contains little water vapor, and 
results because the pressure of the water vapor in the air is below 0.089 psia 
and is therefore in the sublimation range. The thermodynamic principles 
that are involved will be discussed in greater detail in a later chapter. 


THE PURE SUBSTANCE 


121 


7.2. The pvT Surface. The pure substance, in the absence of motion, 
gravity, capillarity, magnetism, and electricity, has only two independent 
properties. Based upon known values of these two, a third property may 
be established that, in combination with the first two, will represent a 
point in a space which has the three properties as its coordinates. Pres¬ 
sure and specific volume are two of the primary properties and are inde¬ 
pendent of each other. It is logical to select the third of the primary 
properties as the dependent property for the purpose of plotting a series 
of points and so developing a surface that will represent the equilibrium 
states of the substance. In that case the resulting surface would be 
designated as the pvT surfaced 



Fig. 7:1. The pvT surface for water (not to scale). 

Figure 7:1 represents the pvT surface for the pure substance. It pic¬ 
tures the form of that surface for water in certain of its details but is not 
drawn to scale and is representative of the general characteristics of the 
surface for all pure substances. Referring to the notation of the diagram, 
the substance is at its triple point at any place along the line abc, a line of 
constant temperature and constant pressure corresponding to the values 
of these properties at the triple point. Points along this line correspond 
to various proportions of the three phases that can be in equilibrium at 
triple-point temperature and pressure. Point a represents (for water) 
the liquid phase alone and point c the vapor phase; point b might represent 
the solid phase alone, although it could also show the volume of a mixture 

1 Any three properties may be used for the delineation of a surface. For instance, 
when the internal energy, the absolute temperature, and the entropy are the space 
coordinates, the surface is called the uTs, or Gibbs, surface and finds use in discussions 
of relative stability and availability. 































122 


BASIC ENGINEERING THERMODYNAMICS 


of liquid and vapor or of all three phases. This possibility will be exam¬ 
ined in more detail in a later article. 1 

The line de is a line of constant temperature and pressure and shows the 
increase of volume as the water changes to ice at that pressure. For a 
substance that becomes less dense as it melts, point d, representing the 
liquid, would be to the right of e. The line eb is the locus of the increasing 
fusion temperatures of ice as the pressure is lowered; volumes along this 
line are those of the solid. Line da plots this same temperature increase 
for the liquid. For most substances the fusion temperature increases 
with increased pressure. The (concealed) surface deba is slightly curved 
toward increased temperatures at lower pressures for water but is per¬ 
pendicular to the pT plane; points on this surface, such as x", represent 
mixtures of the solid and liquid phases. For substances which expand as 
they melt this surface would not be concealed in the view shown in Fig. 
7:1, and the curvature would be in the opposite direction, or toward 
decreased temperature at lower pressure. 

The line aimg is the locus of the possible conditions of the saturated 
liquid, g being the critical point above which equilibrium mixtures of the 
liquid and vapor cannot exist. Similarly, the line cjng describes the locus 
of saturated vapor states. Lines mn and ij, crossing the space between 
these two curves, are lines of constant pressure and temperature and 
therefore normal to the pT plane. Points on this surface such as point x 
represent liquid-vapor mixtures in various proportions according to their 
locations; for example, point x may be estimated by its position to indi¬ 
cate a mixture of about 80 per cent vapor and 20 per cent liquid at the 
pressure and temperature corresponding to i (or j). All possible liquid- 
vapor equilibrium states will lie on the surface outlined by the points 
aimgnjca. This surface curves upward to higher pressures at higher 
temperatures but is everywhere perpendicular to the pT plane. 

Lines byu and czv are both plotted as the locus of sublimation pressures 
and sublimation temperatures, the first at the corresponding specific vol¬ 
ume for the solid, the second for the vapor with which it is in equilibrium. 
Points along the constant-pressure constant-temperature line yz represent 
the volume of mixtures of the solid and the vapor that are in equilibrium 
at this pressure and temperature; for instance, the position of point x' 
indicates a proportion of about three-fourths vapor, one-fourth solid. 
The surface uybczvu, normal to the pT plane, includes equilibrium states 
for the solid-vapor mixture. 

1 The specific volume of water is less than that of ice. For most other substances 
the specific volume of the solid is less than that of the liquid, and point a would fall 
to the right of b. In that case, the position of point b on the surface could indicate 
the volume of unit weight of only the solid phase at the triple point while a could 
represent either all liquid or a mixture of phases. 


THE PURE SUBSTANCE 


123 


As shown on the diagram, the phase borders are definite and distinct 
with the exception of that between the liquid and the gas near the critical 
point. The critical point g lies on both the saturated-liquid line and the 
saturated-vapor line. As the liquid passes this point, no change in any 
property is observed except that the fluid loses that property which is 
characteristic of a liquid and which we call surface tension. It will be 
remembered that surface tension enables a liquid to maintain a distinct 
boundary surface and, at points of contact with the vapor, to form a 
meniscus. The transition across point g has been observed experimen¬ 
tally. The only visual evidence of that transition is the disappearance of 
the meniscus. At this point the liquid does not “boil,” in the sense that 
the change to the vapor phase is gradual; yet there is no abrupt change in 
volume, for the volumes of the liquid and vapor phases are equal. The 
change in phase is instantaneous, and mixtures of the two phases are not 
observed. 

The line pgq is a line of constant temperature corresponding to the 
critical temperature. A point on the surface dhlpgmiad may be definitely 
classified as representing an equilibrium state of the liquid, and the line pg 
is accordingly often taken as being the boundary between the liquid and 
gas phases. The problem is not as simple as that, however, for a state 
such as that represented by a, lying not too far above pg, may have 
characteristics which leave us in doubt as to whether it should be classified 
as liquid or gas. If the pressure is lowered without change of tempera¬ 
ture, the path aa' will be followed. At a' the substance has definitely 
assumed the characteristics of a gas; yet at every point during the transi¬ 
tion the substance has remained homogeneous in phase composition, i.e., 
there has been no mixture of phases. If a is considered to be a liquid 
state, then the change to the gas phase has been, like that observed at the 
critical point, abrupt though not explosive. We shall adopt the policy of 
classifying states such as a simply as fluid states and thus avoiding the 
issue. 

Lines Imno and hijk are lines of respectively constant temperature that 
pass successively over the liquid surface, the two-phase liquid-vapor sur¬ 
face, and the gas surface. Because the liquid is compressible to only very 
slight degree, Im and hi are nearly vertical. Line pg, also a constant- 
temperature locus, deviates noticeably from the vertical; as the critical 
temperature is approached, the liquid becomes more compressible. The 
sections mn and ij across the liquid-vapor surface are normal to the pT 
plane, as has been observed previously. Segments no and jk of these 
constant-temperature lines are curves which, when projected on the pv 
plane, are approximate hyperbolas in shape. As we move to the right 
along these lines (toward lower pressures), the projection of the curve 
more and more becomes truly hyperbolic in shape. A reference to Eq. 


124 


BASIC ENGINEERING THERMODYNAMICS 


(1:6), an equation of state for the so-called ideal gas, indicates that lines 
of constant temperature on the pv plane would, for that class of sub¬ 
stances, be true hyperbolas. At pressures that are low relative to the 
critical pressure, the gas phase of any pure substance approaches the charac¬ 
teristics of the ideal gas. 

The line wyzz', also a constant-temperature line, crosses successively 
the surfaces that represent the solid, the solid-vapor mixture, and the 



vapor. The solid is even less susceptible to compression than the liquid, 
and wy is practically a vertical line. Segment yz is normal to the pT 
plane, of course, and zz', because of the relatively low pressures along its 
entire length, projects on the pv plane as almost an exact hyperbola. 

7:3. The Phase Diagram. Referring to Fig. 7:1, it will be noted that 
the surfaces which represent mixtures of two phases are everywhere per¬ 
pendicular to the pT plane and will project as curved lines on that plane. 
The triple point is normal to the pT plane and will project as a point. 
The entire projection forms a pattern of connecting lines and is called a 






THE PURE SUBSTANCE 


125 


phase diagram . Figure 7:2 illustrates a phase diagram for water; 
included are the lines that divide all seven of the solid phases of this pure 
substance. Because of the wide range of pressures that is covered, a 
logarithmic scale has been used for pressure. Within the range of Fig. 
7:1, Fig. 7:2 is lettered to correspond with that figure. 

The various solid phases of ice are distinguished by differences in 
crystalline structure. The lines that separate each phase from its neigh¬ 
bors have the same significance as the lines in the lower part of the figure 



Fig. 7:3. Pressure-volume diagram for water (not to scale). 

in that they designate pressure-temperature relationships under which 
mixtures of the two phases may exist in equilibrium. For instance, the 
common boundary of ice I and ice III is horizontal and extends between 
temperatures of about —30 and — 8°F, indicating that equilibrium mix¬ 
tures of these two phases may exist at only one pressure but over the 
designated range of temperature. The intersections of these lines may be 
classed as additional triple points since they designate the pressure and 
temperature at which three phases may coexist. 

7:4. The pressure-volume diagram is obtained by projecting the lines 
that separate phases and phase mixtures on Fig. 7:1 to their positions on 
the pv plane. Figure 7:3 is a pressure-volume diagram for water on which 
the lower pressures and specific volumes have been greatly exaggerated 















126 


BASIC ENGINEERING THERMODYNAMICS 


in order to prevent congestion at the left and near the bottom of the dia¬ 
gram; the figure is therefore not to scale, but, as a guide, the pressures at 
the triple point and the critical point and the corresponding volumes of 
the saturated liquid, solid, and vapor are shown on the coordinate axes. 
This diagram is also lettered to agree with Fig. 7:1. The triple point 
appears as a line since one of the coordinates is an extensive property. 
The liquid surface is partly concealed by the surface representing the 
compressed solid; it extends to the right from line da. The boundary 
between the liquid surface and the gas surface is somewhat indefinite, as 
explained in a preceding article, but falls approximately along the line pg. 
The increasing resemblance of the isothermals rs, gq, no, cf, and zz' to 

hyperbolas in this projection as the 
pressure decreases below its value 
at the critical point is a significant 
feature of this diagram. 

7:5. The triple point plots as a 
point on the pressure-temperature 
diagram. When one of these in¬ 
tensive properties is used in con¬ 
junction with an extensive prop¬ 
erty, such as volume (see Fig. 7:3), 
it becomes a line. In general, if 
extensive properties are measured 
along both coordinates, the triple 
point defines an area on the result¬ 
ing diagram. 1 As shown in Fig. 

7:3, at the triple point the specific 
volume of the solid is 0.01747 ft 3 , 
of the liquid 0.01602 ft 3 , and of the 
vapor 3303 ft 3 . The corresponding relative values of the enthalpy, re¬ 
ferred to an arbitrary base to be discussed later are, respectively, —143.3, 
0.02, and 1075.8 Btu. The corresponding points are plotted as S, L, and 
V on Fig. 7:4, an enthalpy-volume diagram. The volume of the solid and 
the liquid have been exaggerated on this chart, for clarity. 



Fig. 7:4. The triple point (water). 


Any point on the line joining L and V in Fig. 7:4 (as point a) represents 
a mixture of only the liquid and vapor phases at triple-point pressure and 
temperature. Similarly, point b denotes a mixture of the solid and the 
vapor and point o a mixture of liquid and solid in equilibrium at this pres¬ 
sure and temperature. The proportions of the two constituents in the 
mixture is indicated by the position of the point. For example, the posi¬ 
tion of point a indicates that the weight of vapor in unit weight of the 


1 An exception is the enthalpy-entropy, or Mollier, diagram. 








THE PURE SUBSTANCE 


127 


liquid-vapor mixture is numerically equal to the ratio of the distance 
between L and a to that between L and V. 

Any point (as point d of Fig. 7:4) that lies within the triangular triple¬ 
point area represents a mixture of all three phases. If the respective 
weights of the three phases in unit weight of the mixture are designated 
as Xs, x L , and x v and the known values of the extensive properties used 
to define state d are p i and p 2 , a set of three simultaneous equations may 
be written, as follows: 

Pi = XsPlg + X L pi L + XvPiy ; p 2 = XsP 2 g + X L p 2b + XvP 2 y 

1 = Xs + X L + Xv 

n which pi 3 , p lL , and p lr are, respectively, the known values of the first 
extensive property for 1 lb of the solid, the liquid, and the vapor and p 2a , 
p 2u and p 2v are the corresponding values of the second extensive property. 

7:6. The Critical Point. If water under the critical pressure of 3206 
psia and at a temperature below the critical temperature of 705°F enters a 
transparent tube and receives heat as it flows slowly through the tube, its 
temperature will progressively increase to the critical temperature. As 
it passes this temperature, it becomes a gas and, if further heat is supplied, 
continues to increase in temperature. This process is shown as line lg2 
in Fig. 7:3. The transition from liquid to gas is not readily observable, 
for surface tension is a property that gradually disappears, vanishing 
entirely at the critical point, and no well-defined meniscus will be apparent 
at temperatures close to the critical temperature, but if some substance 
that had a distinctly different color in the gas phase from that which dis¬ 
tinguished the liquid were put through an equivalent process, the change 
in phase at the critical point would be found to be abrupt. 

In order to observe the disappearance of the meniscus, let us assume 
that a mixture of the saturated liquid and the saturated vapor in exactly 
the right proportions is placed in a transparent container and heated. If 
the pressure and temperature originally are well below their values at the 
critical point, a distinct meniscus, or separation surface, will be apparent 
between the liquid and the vapor. As heat is added, the liquid will boil 
and the pressure and temperature will rise along the curve cjng of Fig. 7:2, 
maintaining equilibrium between the two phases. This boiling indicates 
that the proportion by weight of vapor in the mixture is increasing, but 
the meniscus will be observed to remain near the center of the container. 
This reflects the greater susceptibility of the vapor to compression and the 
characteristic of the liquid that causes it to reflect the effects of tempera¬ 
ture rather than pressure in its volume changes. As the critical pressure 
and temperature are neared, the meniscus becomes less distinct and 
finally vanishes as the critical point is reached and passed. The process 
is illustrated as 3^4 in Fig. 7:3. If too large a proportion of the liquid had 


128 


BASIC ENGINEERING THERMODYNAMICS 


been placed in the container originally, the meniscus would have risen 
during heating and would have reached the top before the critical pressure 
had been attained. If too little liquid had been introduced, the meniscus 
would have vanished at the bottom at a pressure less than the critical 
pressure. 

7:7. Tables of Properties. When an equation of state connecting 
three properties of the pure substance is known, the values of other essen¬ 
tial properties that correspond to a given state may be determined. 
Equation (1:6) is an example of the simplest possible equation of state, 
but it applies with adequate accuracy to only a small portion of the gas 
surface of Fig. 7:1. When the range of states in which interest is centered 
falls within the area to which this simple equation applies, we may, as has 
been shown in earlier chapters, calculate without too much difficulty the 
values (or, at least, the relative values) of certain other properties at 
various states of the substance. To cover the entire surface with suffi¬ 
cient accuracy, a series of equations much more complicated than Eq. 
(1:6) is often required, each covering a specific range and matching 
accurately at the borders of its area of application the results obtained for 
points along that border from another equation of state which has been 
designed to apply to the adjacent area. These equations are normally so 
complicated that a great amount of research and calculation has been 
directed toward the preparation and tabulation of continuous solutions 
for the equations of state for certain especially important substances. 
These tabulations give the values of the more essential properties over a 
wide range of states and are called tables of properties. Their form is more 
or less standardized, and they normally are keyed by pressure, tempera¬ 
ture, or both, and give values of the volume, the enthalpy, and the entropy 
corresponding to states within the range to which they apply. The 
internal energy is sometimes included, at least over a part of the tabulated 
range, but may be obtained from the enthalpy, pressure, and volume by a 
simple calculation. One table, keyed by temperature, usually covers the 
saturation range, and a second, keyed by both pressure and temperature, 
is devoted to the gas or superheated vapor. In exceptional cases, addi¬ 
tional tables are available for the compressed liquid or the solid. 

The most complete, careful, and accurate investigations have concerned 
themselves with the properties of water and steam. Many steam tables 
have been published, each successive tabulation reflecting certain refine¬ 
ments in experimental or other method or covering a new range. Widely 
used in engineering calculations at present, “Thermodynamic Properties 
of Steam” by J. H. Keenan and F. G. Keyes 1 is an example. The dis¬ 
cussion to follow will be based on this tabulation. The solution of many 
of the problems included in this text will require that the reader make use 

1 John Wiley & Sons, Inc., New York, 1936. 


THE PURE SUBSTANCE 


129 


of some standard steam table. The solution of the examples is based on 
the Keenan and Keyes tables whereever the properties of water and steam 
are concerned. 


Table 7:1. Saturation: Temperatures* 


Temp., 

°F, 

t 

Pres¬ 

sure, 

psia, 

V 

Sp. volume 

Enthalpy 

Entropy 

Sat. 

liquid 

Vf 

Evap. 

v fg 

Sat. 

vapor 

Vg 

Sat. 

liquid 

h f 

Evap. 

hf g 

Sat. 

vapor 

h { , 

Sat. 

liquid 

Sf 

Evap. 

s fg 

Sat. 

vapor 

s g 

326 

328 

330 

97.52 

100.26 

103.06 

0.01772 

0.01774 

0.01776 

4.521 

4.403 

4.289 

4.538 

4.421 

4.307 

296.52 
298.60 
300.68 

890.2 

888.6 

887.0 

1186.7 
1187.2 

1187.7 

0.4717 
0.4743 
0.4769 

1.1330 
1.1281 
1.1233 

1 .6047 
1 .6024 
1 .6002 


* Abstracted by permission from Table 1, “Thermodynamic Properties of Steam” by J. H. Keenan 
and F. G. Keyes, John Wiley & Sons, Inc., New York, 1936. 


Table 7:2. Saturation: Pressures* 


Pressure, 

psia, 

Temp., 

°F, 

t 

Sp. volume 

Enthalpy 

Entropy 

Internal energy 

Sat. 

Sat. 

Sat. 


Sat. 

Sat. 


Sat. 

Sat. 


Sat. 

V 

liquid 

Vf 

vapor 

Vg 

liquid 

hf 

Evap. 

hfg 

vapor 

hg 

liquid 

Sf 

Evap. 

Sfg 

vapor 

Sg 

liquid 

Uf 

Evap. 

ufg 

vapor 

Ug 

98 

326.35 

0.01772 

4.517 

296.89 

889.9 

1186.8 

0.4721 

1.1322 

1.6043 

296.57 

808.3 

1104.9 

99 

327.08 

0.01773 

4.474 

297.65 

889.4 

1187.0 

0.4731 

1.1304 

1.6035 

297.33 

807.7 

1105.0 

100 

327.81 

0.01774 

4.432 

298.40 

888.8 

1187.2 

0.4740 

1.1286 

1.6026 

298.08 

807.1 

1105.2 

101 

328.53 

0.01775 

4.391 

299.15 

888.2 

1187.4 

0.4750 

1.1268 

1.6018 

298.82 

806.5 

1105.3 

102 

329.25 

0.01775 

4.350 

299.90 

887.6 

1187.5 

0.4759 

1.1251 

1.6010 

299.57 

805.9 

1105.4 

103 

329.96 

0.01776 

4.310 

300.64 

887.1 

1187.7 

0.4768 

1.1234 

1.6002 

300.30 

805.3 

1105.6 


* Abstracted by permission from Table 2, ‘‘Thermodynamic Properties of Steam” by J. H. Keenan 
and F. G. Keyes, John Wiley & Sons, Inc., New York, 1936. 


Tables 1 and 2 of the Keenan and Keyes tables deal with the properties 
of the saturated liquid and the saturated vapor with which it may be in 
pressure-temperature equilibrium. Table 1 is keyed by temperatures 
(from 32 to 705.4°F, the critical temperature) and Table 2 by pressures 
(from a pressure of 0.25 in. Hg, or 0.123 psia, to an upper limit of 3206.2 
psia, the critical pressure). Tables 7:1 and 7:2 are based on these tables. 

It will be observed that Tables 7:1 and 7:2 cover approximately the 
same range of states and illustrate the convenience of two tables although 
either one alone would be adequate. If, for example, we are interested in 
saturated water, saturated steam, or a mixture of the two at a tempera¬ 
ture of, say, 327.2°F, interpolation is much more simple and more rapidly 
performed if Table 1 is used; similarly, if the pressure is known, the use of 
Table 2 is indicated. Another convenience is the inclusion of tabulated 




























































130 


BASIC ENGINEERING THERMODYNAMICS 


values of the internal energy in Table 2. This is not a usual feature ol 
tables of properties because the internal energy may be easily computed 
from the enthalpy, pressure, and volume. Its inclusion requires some 
crowding and the omission of one of the columns of Table 1. 

The tables are, to a considerable extent, self-explanatory. We are 
already familiar with the use of the symbols v, h, s, and u to denote specific 
values of the volume, enthalpy, entropy, and internal energy, respectively. 
The subscript/may be read as “of the saturated liquid,” the symbol g as 
“of the saturated vapor.” When the subscript/</ appears, it denotes the 
difference between the values of that property for the saturated liquid 
and the saturated vapor or the change of the property as the liquid com¬ 
pletely vaporizes at constant temperature and pressure. It may be read 
as “of evaporation”; for example, h/ g is the “enthalpy of evaporation.” 
Specific volumes are tabulated in cubic feet per pound, enthalpy and 
internal energy in Btu per pound, and the entropy in ranks per pound. 

The values of the enthalpy, entropy, and internal energy are relative 
and are the differences between the values of these properties at the given 
state and at some state at which they are arbitrarily assigned a zero value. 
Enthalpy and entropy have a zero value for the saturated liquid at 32°F. 
Pressure and volume are measured in terms of their absolute values, and 
the internal energy of the saturated liquid at 32°F is therefore very 
slightly negative; the difference is so small that it has little practical 
significance. As a matter of technical accuracy, the choice of the satu¬ 
rated liquid at 32°F as the base level is somewhat inconsistent, for this 
temperature is slightly below the triple-point temperature and this sub¬ 
stance cannot exist as a liquid at the combination of temperature and 
pressure as stated in the first line of Table 1 of the unabridged tables; it 
must be a vapor, a solid, or a mixture of these two phases. From the 
practical standpoint, a level that corresponds with the lower anchor point 
of the Fahrenheit scale of temperature has more convenience, and this 
outweighs the very slight inaccuracy that is involved. 

Data taken from Table 1 or Table 2 may be applied to the calculation 
of the properties of an equilibrium mixture of the liquid and vapor phases. 
The pressure and temperature of such mixtures will, of course, correspond 
in accordance with the relationship presented in those tables regardless of 
the proportions of liquid and vapor. The value of any extensive property 
(volume, enthalpy, entropy, internal energy) is the sum of its values for 
the parts. The quality x denotes the weight of vapor in 1 lb of the mix¬ 
ture, and the weight of liquid is thus 1 — x. We may therefore write 

v = (1 — x)v f + xv g 

h = (1 — x)hf + xh 0 

s = (1 — x)s f + XSg 

u = (1 — x)u f + xug 


( 7 : 1 ) 


THE PURE SUBSTANCE 


131 


in which v, h, s, and u are the values of the corresponding properties for 1 
lb of the mixture of quality x. Taking into consideration the fact that 
v o = v f + Vfg, etc., each of these equations may be converted to the follow¬ 
ing two forms, which may be used according to relative convenience: 


V = Vf + XVfg = Vg — (1 — x)Vfg 

h = hf + xhfg = hg - (1 - x)hfg 
S Sf —f- XSfg Sg (1 X^Sfg 
U — Uf T XUfg = Ug — (1 — X)Ufg. 

When Table 7:1 is used and the internal energy cannot be calculated 
directly from tabulated values, it may be obtained as u = h — 144 pv/J. 
The specific volume of the liquid is usually so small as compared with that 
of the vapor that the volume of the former is often neglected, and, as an 
approximation, v = xv g . This approximation may be used only in the 
case of the volume and becomes less accurate at low quality and at high 
temperatures and pressures. 

Example 7:7 A. Find the temperature and values of the specific volume, enthalpy, 
entropy, and internal energy for wet steam at a pressure of 85 psig and having a 
quality of 0.90 (90 per cent). Atmospheric pressure is 30.56 in. Hg. 

Solution. The pressure of the atmosphere is (0.491) (30.56) = 15.0 psia Adding 
this to the gage pressure, the pressure of the steam is 100 psia. The corresponding 
saturation temperature is 327.81°F. This temperature will apply to all qualities 
at this pressure. 

At high qualities the calculation of the values of the extensive properties is more 
easily and accurately made by using the second form of Eqs. (7:2). Thus 

v = Vg - (1 - x)v f0 = 4.432 - (1 - 0.9) (4.432 - 0.018) = 4.432 - 0.4414 = 3.991 

ft 3 /lb 

h = hg - (1 - x)hfg = 1187.2 - (1 - 0.9) (888.8) = 1187.2 - 88.9 = 1098.3 Btu/lb 
8 = s 0 - (1 - x)s /g = 1.6026 - (1- 0.9)(1.1286) = 1.6026 - 0.1129 = 1.4897 
u = u g — (1 — x)u/g = 1105.2 - (1- 0.9) (807.1) = 1105.2 - 80.7 = 1024.5 Btu/lb 

Note that it is not appropriate to carry the calculation to a greater number of signifi¬ 
cant places than is presented in the tables. 

The quality is high and the pressure moderate so that the approximation might 
have been applied in the calculation of specific volume. Checking, 

v = XVg = (0.90) (4.432) = 3.989 ft 3 /lb 

The internal energy may alternately be calculated as 

u = h - 14 t— = 1098.3 - (144)(1 ^ (3,991) = 1098.3 - 73.8 = 1024.5 Btu/lb 
J 778 

Table 3 of the Keenan and Keyes tables’tabulates the properties of 
superheated steam in the form shown in Table 7:3. 

Temperature and pressure are independent properties of the gas or 
superheated vapor and, in that range, may be used to establish all other 
properties of the pure substance. Table 3 is keyed by pressure and tem¬ 
perature and is most convenient for use when these two properties are 




132 


BASIC ENGINEERING THERMODYNAMICS 


known. However, the table may be entered with any two of the tabu¬ 
lated properties (pressure, temperature, volume, enthalpy, entropy), 
and if the assigned values lie within the scope of the tabulation, systematic 
search will locate the corresponding state and, with it, the values of the 
other properties. The range of pressures covered by the table is from 1 
psia to the critical pressure of 3206.2 psia, and the range of temperatures 
is from the saturation temperature corresponding to each pressure to 
1600°F. The last portion of Table 3 in its unabridged form carries the 
pressures above the critical pressure to a maximum of 5500 psia. A part 
of the data in this section of the tabulation applies to the “twilight zone” 
to which reference has previously been made and even overlaps across line 
pg of Figs. 7:1 and 7:3 to encroach on the compressed-liquid surface. 


Table 7 : 3 . Superheated Vapor* 


Pressure, psia 
(sat temp.) 


Sat. 

liquid 

Sat. 

vapor 

Temperature, °F 

330 

340 

350 

360 

370 

380 

390 

400 

420 

99 

V 

0.018 

4.474 

4.496 

4.569 

4.641 

4.712 

4.783 

4.852 

4.921 

4.989 

5.124 

(327.08) 

h 

297.6 

1187.0 

1187.7 

1194.6 

1200.3 

1205.9 

1211.5 

1217.0 

1222.4 

1227.7 

1238.3 


s 

0.4731 

1.6035 

1.6056 

1.6130 

1.6201 

1.6271 

1.6338 

1.6403 

1.6467 

1.6530 

1.6651 

100 

V 

0.018 

4.432 

4.448 

4.521 

4.592 

4.663 

4.732 

4.801 

4.870 

4.937 

5.071 

(327.81) 

h 

298.4 

1187.2 

1188.5 

1194.3 

1200.1 

1205.7 

1211.3 

1216.8 

1222.2 

1227.6 

1238.1 


s 

0.4740 

1.6026 

1.6043 

1.6117 

1.6188 

1.6253 

1.6325 

1.6391 

1.6455 

1.6518 

1.6639 

102 

V 

0.018 

4.350 

4.355 

4.427 

4.497 

4.567 

4.635 

4.703 

4.770 

4.836 

4.968 

(329.25) 

h 

299.9 

1187.5 

1188.0 

1193.9 

1199.7 

1205.3 

1210.9 

1216.4 

1221.9 

1227.3 

1237.9 


s 

0.4759 

1.6010 

1.6016 

1.6090 

1.6162 

1.6232 

1.6300 

1.6366 

1.6430 

1.6493 

1.6615 


* Abstracted by permission from Table 3, “Thermodynamic Properties of Steam” by J. H. Keenan 
and F. G. Keyes, John Wiley & Sons, Inc., New York, 1936. 


The units in which the various properties are measured are the same as 
in Tables 1 and 2. The internal energy is not tabulated but may be 
readily calculated as u = h — 144 pv/J. Occasionally it is necessary to 
identify a state when one of the two known properties is the internal 
energy. Table 3 does not lend itself readily to this purpose but may be 
used by first locating the approximate area in the table that includes the 
required state by methods which will suggest themselves and then apply¬ 
ing trial-and-error or graphical methods to locate the state more accu¬ 
rately. It will be noted that, grouped with the pressure along the left 
border of Table 3, data taken from Table 2 and covering the properties of 
the saturated liquid and vapor are tabulated. This will often be a con¬ 
venience as when the calculation of the increase of a property during 
superheating at constant pressure is of interest. 



































THE PURE SUBSTANCE 


133 


Example 7:7 B. Determine the specific volume, enthalpy, entropy, and internal 
energy of steam at a pressure of 100 psia and a temperature of 400°F. What is the 
mean specific heat at constant pressure? Compare it with the instantaneous value at 
400°F. Refer to Fig. 6 of the Keenan and Keyes tables, and check. 

Solution. The specific volume, enthalpy, and entropy are read directly from Table 
3 and are, respectively, 4.937 ft 3 /lb, 1227.6 Btu/lb, and 1.6518. The internal energy 
may be calculated as 

u = h — _ 1227.6 - ( 144) (1 00) (4 . 9 3 7) = 1227 f . _ gi 3 = n363 Btu/lb 

The mean, or average, specific heat at constant pressure is (A h/At) p . Starting 
from the saturated vapor at 100 psia, the increase of specific enthalpy has been 
1227.6 — 1187.2 = 40.4 Btu and the increase of temperature 400 — 327.81 = 72.19°F. 
The mean specific heat at constant pressure, c Pm = (40.4/72.19) = 0.56 Btu/(°F) 
(lb). The instantaneous specific heat, c Pi , may be determined with sufficient 
accuracy by calculating ( Ah/At) p over a range of temperatures for which 400°F is the 
mid-point. Table 3 makes it convenient to choose this interval as 380 to 420°F, and 
the change of temperature is therefore 40°F. The corresponding change of specific 
enthalpy is 1238.1 - 1216.8 = 21.3 Btu and c Pi = 21.3/40 = 0.532 Btu/(°F) (lb). 
It is the instantaneous value of the specific heat that is plotted in Fig. 6 of the Keenan 
and Keyes tables, and our value, when spotted on that figure, seems in good agreement 
with the data which it presents. 


Table 7:4. Compressed Liquid* 


Pressure, 

psia, 

(Sat. temp.) 


Temp., °F 

32 

100 

200 

300 

400 

500 

600 

Sat. p 

liquid vf 

hf 

Sf 

0.08854 
0.016022 

0 

0 

0.9492 
0.016132 
67.97 
0.12948 

11.526 

0.016634 

167.99 

0.29382 

67.013 

0.017449 

269.59 

0.43694 

247.31 

0.018639 

374.97 

0.56638 

680.8 

0.020432 
487.82 
0.68871 

1542.9 

0.023629 

617.0 

0.8131 

1000 

(v - Vf) X 105 

-5.7 

-5.1 

-5.4 

-6.9 

-8.7 

-6.4 


(544.61) 

h — hf 

+2.99 

+2.70 

+2.21 

+ 1.75 

+0.84 

-0.14 



(s - Sf) X 103 

+ 0. 15 

-0.53 

-1.20 

-1.64 

-2.00 

-1.41 


2000 

(v — vf) X 10 5 

-11.0 

-9.9 

-10.8 

-13.8 

-19.5 

-27.8 

-32.6 

(635.82) 

h — hf 

+ 5.97 

+ 5.31 

+4.51 

+3.64 

+ 2.03 

-0.38 

-2.5 


(s - Sf) X 103 

+0.22 

-1.18 

-2.39 

-3.42 

-4.57 

-5.58 

-4.3 


* Abstracted by permission from Table 4, ‘‘Thermodynamic Properties of Steam” by J. H. Keenan 
and F. G. Keyes, John Wiley & Sons, Inc., New York, 1936. 


The properties of the compressed liquid are covered in Table 4 up to a 
pressure of 6000 psia. The liquid surface of Fig. 7:1 is covered up to that 
pressure with the exception of a small triangular area below 32°F, where 
the fusion temperature is slightly below 32°F because of the imposed pres¬ 
sure; the omission of this area is unimportant. Sample lines based on 
Table 4 are shown in Table 7:4. 

Table 4 in its unabridged form carries the temperatures up to the 
































134 


BASIC ENGINEERING THERMODYNAMICS 


critical temperature of 705.4°F. Grouped with these temperatures at 
the top of the table are the corresponding saturation pressure and the 
values of the specific volume, enthalpy, and entropy for the saturated 
liquid. The increments of these properties, as the result of the applica¬ 
tion of a superpressure on the surface of the liquid, are small, and their 
values are accordingly stated, when the available information warrants 
that practice, to a larger number of significant places than in Table 1. In 
the lower part of the table, the increments of each of these properties that 
correspond to each superpressure are tabulated. These increments are 
applied as additions to or subtractions from the corresponding values for 
the saturated liquid and are arranged, for convenience, in a manner such 
that the last place of the increment corresponds with the last place shown 
in the tabulation at the head of the column. Thus at a pressure of 2000 
psia and a temperature of 100°F the specific entropy of the compressed 
liquid is 0.12948 - 0.00118 = 0.12830, while at the same pressure but at 
a temperature of 600°F the corresponding calculation is 0.8131 — 0.0043 
= 0.8088. Some time may be saved in the use of Table 4 by remember¬ 
ing and applying this feature. 

A detailed examination of Table 4 reveals that the degree of compressi¬ 
bility of water increases at an accelerated rate as the critical temperature 
is approached. At 5000 psia the volume is less than the volume of the 
saturated liquid by less than 2 per cent at 32°F and by about 46 per cent 
at the critical temperature. Not so very many years ago very little 
quantitative information was generally available with respect to the 
compressibility of water, and it was customary to assume it to be incom¬ 
pressible in calculating the properties of the superpressure liquid. On 
the basis of that inaccurate assumption, if pressure were applied on the 
surface of the saturated liquid (with no heat flow), there would be no 
change in volume and thus no work would be in evidence and no energy 
would cross the boundaries of the liquid system. As a consequence, 
neither the internal energy nor the entropy could change, and the enthalpy 
of the superpressure liquid was necessarily greater than that of the satu¬ 
rated liquid at the same temperature. When compressibility is taken into 
account, the decrease in volume represents energy in the form of work that 
enters the system. As a result of the entry of this energy, the tempera¬ 
ture of the liquid rises slightly, and, in order to return it to the reference 
level of temperature, heat must be taken from the system. This negative 
heat flow is accompanied by a decrease of entropy, as shown over the 
major portion of Table 4, and is also accompanied by a decrease of internal 
energy. When the degree of compressibility is small, as for temperatures 
up to about 400°F, the effect of this decrease of internal energy on the 
enthalpy is not large enough to cancel the increase of enthalpy that is 
caused by the higher pressure and the increment of enthalpy, with respect 


THE PURE SUBSTANCE 


135 


to the enthalpy of the saturated liquid, is positive. At higher tempera¬ 
tures and correspondingly higher compressibility, this increment becomes 
negative. At 32°F it is noted that the increment of entropy is positive; 
this is due to the unusual characteristic of liquid water in having its 
greatest density not at the fusion temperature but a few degrees above 
that point. 

Example 7:7 C. A pump receives saturated water at a temperature of 200°F and 
discharges it continuously at a pressure of 1500 psia. If the compression is isentropic 
(adiabatic and reversible), calculate (a) the final temperature of the water and (6) 
the work required per pound of water pumped. 

Solution: 

(a) The state of the water at entrance will be designated by the subscript 1, that at 
exit from the pump as 2. Then, using Table 4, p x = 11.526 psia; v x = 0.016634 ft 3 /lb; 
h\ = 167.99 Btu/lb; Si = s 2 = 0.29382. At p 2 = 1500 psia, the specific entropy of 
the compressed liquid is 0.29203 at 200°F, 0.43441 at 300°F. Interpolating between 
these limits, the temperature corresponding to s 2 is found to be 201.25°F. An inter¬ 
polation over so wide a range is, of course, subject to some criticism, and the final 
temperature might differ from our result by a few hundredths of a degree. 

( b ) At 1500 psia and 200°F, the specific enthalpy of compressed water is 167.99 + 3.36 
= 171.35 Btu; at the same pressure and 300°F, it is 272.29 Btu. Interpolating for a 
temperature of 201.25°F, h 2 is found to be 172.61 Btu. Applying the steady-flow 
energy equation [Eq. (3:5)] and neglecting differences in velocity and elevation at 
entrance and exit, iW 2 /J = hi — h 2 = 167.99 — 172.61 = —4.62 Btu/lb. 

Figure 3, located just below Table 4 in the Keenan and Keyes tables, is a con¬ 
venience for the solution of problems of this character. Figure 3 is similar to a 
Mollier chart for the compressed liquid except that the difference between the enthal¬ 
pies of the compressed and the saturated liquids is measured along the ordinate 
instead of the total enthalpy. Applying it to the solution of this problem, point 1 
is located at the intersection of the line of constant temperature of 200°F with the 
bottom line of the diagram, which represents saturation states. Projecting vertically 
upward (at constant entropy) to the line representing the final pressure of 1500 psia, 
the results of our calculation are very closely checked in a fraction of the time required 
to carry out the detailed computation that is made above. 

For the isentropic process, dh = v dP/J (see Example 3:6), and since the change in 
volume is minor, this provides an easy, if approximate, method of calculating the 
change of enthalpy as 

7 vAP 0.016634(1500 - 11.5)(144) . 

AA = ~T = -778- = 4 ' 57 BtU 

Table 5 presents the properties of the solid and the vapor in the upper 
portion of the temperature range in which a mixture of these two phases 
may be in temperature-pressure equilibrium. It is in much the same form 
as Table 1, being keyed by temperature. A few sample lines are pre¬ 
sented in Table 7:5. 

Table 5 requires little detailed explanation. The range of the una¬ 
bridged table is from 32 down to — 40°F. At the latter temperature the 
saturation pressure is 0.0019 psia. The subscript g has the same meaning 



136 


BASIC ENGINEERING THERMODYNAMICS 



* Abstracted by permission from Table 5, “Thermodynamic Properties of Steam” by J. H. Keenan and F. G. Keyes, John Wiley & Sons, Inc., New York, 1936. 

































THE PURE SUBSTANCE 


137 


as before. The subscript i may be read as “of the saturated solid/’ and 
ig refers to the change of the property during sublimation. The same 
reference levels for the evaluation of enthalpy, entropy, and internal 
energy are, of course, retained. 

Also included in the Keenan and Keyes tables are Table 6, giving the 
viscosity of saturated water and both saturated and superheated steam, 



and Table 7, which tabulates the thermal conductivity of the saturated 
and superheated vapor. 

7:8. The Temperature-Entropy Diagram. Figure 7:5 shows a tem¬ 
perature-specific entropy diagram for water. Saturated-liquid states 
have been plotted from steam-table values as the line ag, the saturated- 
vapor states as gcz. The line by is the locus of the states of the saturated 
solid. Between ag and gc lies the area that represents equilibrium mix¬ 
tures of the liquid and the vapor, and mixtures of the solid and the vapor 
are located in the area below the triple-point line and between yb and 
cz. The superheated-vapor area lies to the right of gz and extends into 
the region above g. The compressed-liquid area lies close alongside the 












138 


BASIC ENGINEERING THERMODYNAMICS 


line ag and to its left except below the temperature ol greatest density of 
water (39°F), where it crosses to the right side of this line. The com¬ 
pressed-solid states, for which no quantitative data are available to us in 
the tables, lie closely along the left side of yb. 

Lines of constant pressure of 100, 14.7, and 5 psia have been drawn 
across the liquid-vapor area of Fig. 7:5 and extended into the superheated- 
vapor region. If the chart were extended downward to a temperature of 
absolute zero ( —460°F), the area under these lines, as was brought out in 
Chap. 6, would be equal to the change of enthalpy between the states 
which they connect. Thus the area under mn would be hf g at 100 psia, 
or 888.8 Btu. This area is rectangular and easily measured as the product 
of its height, or the saturation temperature in degrees Rankine, and its 
width, the change of entropy during evaporation, S/ g ; thus hf g = 
T s;i tSf g . Similarly, the area under a line of constant pressure in the 
superheated-vapor region will measure changes of enthalpy between 

points on that line. This area is less readily calculated but equals, of 
r Ti 

course, / c v dT, in which c v is variable and depends on both the tem- 

perature and the pressure of the vapor. This variation is shown in Figs. 
6 and 7 of the Keenan and Keyes tables. If a constant value (equal to 
the mean, or average, value over the temperature range) is assigned to c p , 
the change of entropy may, conversely, be calculated as in Chap. 6. The 
result will be an approximation. 

A much more detailed and complete temperature-entropy chart is 
included as Fig. 9 of the Keenan and Keyes tables. This chart shows not 
only lines of constant pressure but also lines of constant enthalpy, con¬ 
stant superheat, constant moisture (quality), and constant volume. The 
areas below the lines of constant volume, as explained in Chap. 6, are the 
changes of internal energy between states that they connect when that 

area is measured down to a temperature of absolute zero. 

« 

Example 7:8. Check, approximately, the tabulated values of s/ and Sf 0 at 100 psia 
and the tabulated value of s at 100 psia and 400°F. 

Solution: Referring to Eq. (6:8), ds = 4- The change of volume of the 

liquid is minor, and the second term may accordingly be dropped for the purpose of 
our approximate calculation. At 100 psia, u/ is 298.08 Btu, and, at 32°F, u/ is so 
slightly less than zero that it may be assumed to have that value. By definition, 

_ Idu\ , _ Au _ 298.08 

C * and Crm At 327.81 — 32 — 1,01 BtU// ( F )( lb ) 

Then 

[T dT f 787.8 dT 787 8 

*/ = J To ~f = 1-01 J 492 = 1-01 log, ^ = (1.01)(0.470) = 0.474 

This checks the tabulated value in spite of the approximations introduced by dropping 
the second term of Eq. (6:8) and using c v as a constant instead of a variable. 




THE PURE SUBSTANCE 


139 


To calculate the entropy of evaporation, s fg = = |§§4 = 1.129. This is, in 

1 sat < 87.8 

principle, an exact calculation. The use of the slide rule and of an approximate 
value for the absolute temperature of saturation account for the small discrepancy 
when compared with the tabulated value. 

In checking the tabulated value of s at 100 psia and 400°F, it is noted that the 
enthalpy of superheat (the increase of enthalpy over that of the saturated vapor at 
the same pressure) is 1227.6 - 1187.2 = 40.4 Btu and that the corresponding tempera¬ 
ture rise is 400 — 327.81 = 72.19°F. The mean specific heat at constant pressure 
is therefore 

/Ah\ _ J0A 

Pm )v 72.19 0,56 

Then 

[T dT f 860 dT 860 

s - s a = Cp m J t t ~Y = °-56 J 787 8 ~y = 0.56 log e ^y-g = (0.56)(0.088) = 0.0493 

Adding this entropy of superheat to s g , s = 0.0493 + 1.6026 = 1.6519. This checks 
the tabulated value closely, although an approximation has again been introduced in 
using an average value of c p as a constant value over the entire range. As the range 
of temperature increased, the accuracy of this method of calculation would decrease. 
The tabulated values take into account this variation of specific heat and have been 
obtained as the result of much more careful computation than is possible in the use 
of the slide rule. 

Lines of constant pressure corresponding to 100, 14.7, and 5 psia have 
not been shown for the compressed liquid in Fig. 7:5. These lines, as they 
move to lower temperatures, would lie so close to the saturated-liquid line 
that to attempt to show them on the diagram would be confusing rather 
than helpful. A constant-pressure line at 3206.2 psia, the critical pres¬ 
sure, has been shown, however, to indicate the general form of these lines 
in the compressed-liquid region; even for this line the divergence from the 
saturated-liquid line has been somewhat exaggerated. At 39°F, the 
temperature of greatest density of liquid water, it cuts across the satu¬ 
rated-liquid line and enters the liquid-vapor space of Fig. 7:5. Under 
this pressure the fusion temperature is lowered somewhat, as is indicated 
in the figure. 

Lines of constant pressure lower than the critical pressure will, in the 
compressed-liquid space, lie between the critical-pressure isobar and the 
saturated-liquid line, intersecting both lines at a temperature of 39°F. 
Isobars for pressures above the critical pressure will fall to the left of the 
critical-pressure isobar but very close to it above the temperature of 39°F 
and will also intersect the saturated-liquid line at this temperature. 

The paragraphs immediately above have indicated an overlapping of 
the compressed-liquid space into the liquid-vapor region on the Ts dia¬ 
gram. This is an unusual characteristic of water and is due to its attain¬ 
ment of a maximum density of the liquid at a temperature that is above 
the fusion temperature and to its expansion as the temperature is lowered 






140 


BASIC ENGINEERING THERMODYNAMICS 


below 39°F. It provides an exception to the statement made earlier in 
these pages that, if two independent properties of the simple system, 
composed of a pure substance, are known, all other properties are fixed. 
Ice, water, and steam, or any combination of these phases, meet the 
specifications for a pure substance as set forth in Chap. 1. Temperature 
and specific entropy would constitute two independent properties. As 
obtained from Table 4 of the Keenan and Keyes tables, the specific 
entropy of compressed water at 32°F and 3206.2 psia is +0.00029, and 
its volume is slightly less than the volume of the saturated liquid at the 
same temperature. But the mixture of saturated liquid and saturated 
vapor at 32°F which has a quality of 0.00029/2.1877, or about 0.013 per 
cent, would also have the same entropy and temperature and its volume 
would obviously be greater than that of the saturated liquid. The area 
of overlap is small; that fact and the almost unique character of this 
exception to the general rule combine to make it relatively unimportant 
from a practical standpoint in engineering thermodynamics, but the 
special characteristics of water such as that mentioned in the above lines, 
the lesser density of the solid phase as compared with the liquid, and the 
reduction of fusion temperature with increased pressure are very impor¬ 
tant in ameliorating the conditions under which life exists on the earth’s 
surface. That water, a substance so important to the life process, should 
vary from the general rule certainly is significant. 

7:9. The Mollier diagram often offers great convenience in the solution 
of the practical engineering problem, especially when steady-flow proc¬ 
esses are involved (see Chap. 6). In Fig. 7:6 is shown a Mollier, or 
enthalpy-entropy, diagram for water; the notation corresponds with 
Fig. 7:5. 

The saturated liquid line amg continues to form the saturated-vapor line 
gncz; the critical point g is an inflection point on this saturation line. The 
triple-point line extends from b through a to c (note that the triple point is 
represented by a straight line on this diagram although its coordinates are 
both extensive properties), b representing the saturated solid, a the satu¬ 
rated liquid, and c the saturated vapor. The space under the saturation 
line and above the triple-point line represents mixtures of the liquid and 
the vapor; that below the triple-point line is the solid-vapor area. Above 
the saturated-vapor line lie superheated-vapor states, and the compressed- 
liquid region lies just to the left of and closely borders the saturated-liquid 
line. 

Lines representing constant pressures of 100, 14.7, and 5 psia are shown 
on Fig. 7:6 crossing the liquid-vapor space and extending into the super¬ 
heated-vapor area above. In the liquid-vapor space these are also lines 
of constant temperature of, respectively, 327.8, 212, and 162.2°F. It has 
been shown in Chap. 6 that the slope of a constant-pressure line on the 


THE PURE SUBSTANCE 


141 


Mollier diagram is equal to its absolute temperature, and these lines, as 
they cross the liquid-vapor area, are accordingly straight and diverge as 
they move Irom the saturated-liquid line to the saturated vapor line. At 
the point where they contact the saturated-liquid line, their pressure and 
temperature are the same as (instantaneously) for the saturated liquid, and 
these lines therefore leave the saturated-liquid line as tangents to that 
curve. As they move across the saturated-vapor line into the super- 
heated-vapor region, the temperature, and consequently their slope, 
increases and they curve gradually upward in this area. 



Fig. 7:6. Mollier diagram for water. 

At the saturated-vapor line, the lines of constant temperature diverge 
from those of constant pressure and curve upward to the right. As they 
near the right-hand border of the figure, they become almost horizontal 
(constant-enthalpy) lines. Particularly is this true at the lower tempera¬ 
tures and pressures where the lines of constant temperature leave the 
saturated-vapor line as nearly horizontal lines. We shall later have 
occasion to recall this feature to the mind of the reader. 

A reference to Eq. (7:2) will indicate that the change of enthalpy during 
evaporation is directly proportional to the change of quality of the mix¬ 
ture if the pressure is constant. Thus the intercept of a line of constant 
pressure between the saturated-liquid and saturated-vapor lines may be 
divided into, say, 10 equal parts, and the points of division between these 
parts may be designated as denoting states having qualities of 10, 20, and 




142 


BASIC ENGINEERING THERMODYNAMICS 


so on, up to 90 per cent. On Fig. 7:6 the locus of the 90 per cent quality 
points has been connected and is a line of constant quality. 

The practical use of the Mollier diagram is usually limited to the super¬ 
heat region and the higher qualities. A large-scale Mollier chart, which 
covers the area indicated in Fig. 7:6, accompanies the Keenan and Keyes 
tables. The information it carries is, of course, much more complete and 
detailed than that shown in Fig. 7:6. 

Example 7:9. Steam enters a reversible adiabatic turbine at 200 psia and 500°F 
and leaves at atmospheric pressure (standard). Assuming steady flow and neglecting 
differences in velocity and elevation between entrance and exit, what work is delivered 
per pound of steam flow? 

Solution. The work per pound of flow is hi — hi (see Chap. 3). The expansion 
that takes place in the turbine is indicated as isentropic in the statement of the 
problem, and thus s 2 = Si. Referring to Table 3 of the Keenan and Keyes tables, 
hi = 1268.9 Btu, and si = 1.6240 = s 2 . This value of 1.6240 lies between s/ and s 0 
at the exhaust pressure of 14.7 psia, and the exhaust from the turbine is therefore 
in the form of wet steam. Applying Eq. (7:2), s / 2 + xs/ 02 = Si or 0.3120 + 1.4446a; 
= 1.6240 and, solving, x = 0.908. Then hi = 1150.4 — (1 — 0.908) (970.3) = 
1061.2 Btu. The work of this ideal turbine is 1268.9 — 1061.2 = 207.7 Btu per 
pound of steam. 

The use of the Mollier chart makes the solution of this problem much simpler and 
more rapid. Finding the intersection of a line representing a constant temperature 
of 500°F with the line of 200 psia pressure, hi is read at the side of the chart as 1269 
Btu. Projecting vertically downward (at constant entropy) from this point, the 
intersection with the line representing the exhaust pressure of 14.7 psia falls at an 
enthalpy (hi) of 1061 Btu and a moisture percentage of 9.2, equivalent to a quality of 
90.8 per cent. It is observed that the accuracy in the use of the chart is at least 
within 1 Btu, which is sufficient for engineering uses. Even when greater accuracy 
is desired, the chart may be used to good effect as a check on detailed computations. 

7:10. The Experimental Determination of Quality. The properties 
most readily and accurately determined by observation and experiment 
are the pressure and temperature. For superheated steam and com¬ 
pressed water these are sufficient to establish the state of the fluid; for the 
liquid-vapor mixture these two properties are interdependent and do not 
suffice to locate the proportions of the mixture. It is often necessary to 
determine the quality of the steam as it leaves a boiler or enters a prime 
mover. This can, of course, be done by making use of the greater density 
of the liquid to separate it from the mixture and then comparing its weight 
with that of the steam in which it was carried. The weight of the steam 
portion is usually measured by allowing it to pass through a calibrated 
orifice. A device of this kind is called a separating calorimeter. 

Another method of determining the quality of wet steam is through the 
use of a throttling calorimeter. This device, illustrated in Fig. 7:7, 
includes a sampling tube B, pierced by a number of holes so located that a 
representative steam sample will be obtained and inserted into the steam 


THE PURE SUBSTANCE 


143 


line, an orifice 0 through which the steam sample flows continuously into 
a chamber C, from which it flows directly into the atmosphere. Provision 
is also made, as shown, for determining the pressure of the steam in the 
line and its pressure and temperature in the calorimeter chamber. The 
entire calorimeter is covered by a thick layer of heat-insulating material 
so that the flow of steam through it may be considered adiabatic in charac¬ 
ter. The differences in velocity and elevation of the steam in the line 
and in the chamber may be ignored with negligible error, and no external 
work is performed. Applying the steady-flow energy equation of Chap. 
3, it may be shown that the enthalpies in the line and in the calorimeter 
chamber are equal. 

The use of the throttling calorimeter for the determination of quality 
is limited to situations where the quality of the steam in the line is high 
(about 95 per cent or more) and to 
steam-line pressures in the approxi¬ 
mate range of 75 to 1200 psia, de¬ 
pending on the quality. The pur¬ 
pose of the throttling expansion is 
to change the steam sample to 
superheated steam so that the pres¬ 
sure and temperature in the calo¬ 
rimeter chamber will be sufficient to 
establish its enthalpy. This en¬ 
thalpy, equal to that in the steam 
line, is then used in combination 
with line pressure to locate the 
quality in the line. The solution is 
conveniently obtained graphically 
by making use of the Mollier di¬ 
agram. In this case, the point on 
the chart that represents the pres¬ 
sure and temperature of the super¬ 
heated steam in the calorimeter chamber is located and projected horizon¬ 
tally (at constant enthalpy) to the left across the saturation line to the line 
of constant pressure corresponding to the pressure in the line. The quality 
in the line is read from the chart at this intersection. The throttling 
process does not proceed, however, at constant enthalpy; it merely begins 
and ends at states having equal enthalpies. As the steam passes through 
the orifice, there is a temporary increase of velocity at the expense of a 
decrease of enthalpy; as this velocity is dissipated in the chamber, the 
enthalpy again builds up to its upstream value. 

The reason for the limitation in the use of the throttling calorimeter to 
the range of qualities and pressures mentioned above is made clear by 









































144 


BASIC ENGINEERING THERMODYNAMICS 


Fig. 7:8. This diagram plots the saturation temperatures of the satu¬ 
rated liquid and the saturated vapor as ordinates against their specific 
enthalpies as abscissas. A sample condition for the wet steam in the line 
is spotted on the diagram as the point L; by estimating its relative dis¬ 
tance from the saturated-liquid and saturated-steam lines, the correspond¬ 
ing quality of the steam in the line is seen to be about 9/ per cent. The 
point C indicates the superheated state that is reached as the result of the 

t 



throttling expansion. The line connecting L and C is vertical (at con¬ 
stant enthalpy) but is dotted to remind the reader that points along it do 
not correspond to intermediate states during the expansion. 

The point C must lie in the superheated-steam region if the throttling 
calorimeter is to be usable for the determination of the quality in the line. 
It is apparent that if that quality is too low or line pressure too high, point 
C will lie either in the liquid-vapor region or so close to the saturated- 
steam line that the temperature at C will be very little above the satura¬ 
tion temperature at atmospheric pressure. A superheat of 5 to 10°F is 
usually considered desirable to make certain that the steam in the cham¬ 
ber is superheated; this also accounts for the lower limit of line pressure 








THE PURE SUBSTANCE 


145 


which is shown on the diagram. The pressure and quality ranges can be 
somewhat increased by making provision for calorimeter exhaust into a 
space in which the pressure is less than atmospheric, as to a condenser. 

Example 7:10. T.he pressure in a steam line is 200 psia. In the chamber of a 
throttling calorimeter, the temperature of a sampling of steam from this line has 
become 260°F at atmospheric pressure (standard). What is the quality of the steam 
in the line? 

Solution. The state of the steam in the line will be designated by the subscript 1, 
that in the calorimeter as 2. Then, consulting Table 3, at p 2 = 14.7 psia, t 2 = 260°F, 
h 2 is found to be 1173.8 Btu. This is also the value of hi and 

h fi + x t h f0i = h 2 or 355.36 + 843.0z = 1173.8 

Solving, 

xi = 0.971 

The Mollier chart is again well adapted to the solution of the throttling-calorimeter 
problem. Locating the point on the chart that represents the condition in the calorim¬ 
eter chamber as the intersection of the lines of 14.7 psia pressure and 260°F tempera¬ 
ture in the superheat region above the saturation line, we need not stop to identify 
the enthalpy but can project horizontally (at constant enthalpy) to the left across the 
saturation line until the line which represents line pressure of 200 psia is intersected. 
At that intersection we read a moisture percentage of 2.9, corresponding to a quality of 
97.1 per cent. 

Problems 

1. Based on Figure 7:2, how many triple points may be observed for the pure sub¬ 
stance H 2 0? For each, list the approximate temperature and pressure and the com¬ 
ponents of the mixture. 

2. Is it possible for ice I to melt when the temperature of the ice is above 32°F? 
When its temperature is above 32.02°F? 

3. Will a triple point appear as a point, as a line, or as an area on each of the follow¬ 
ing diagrams? (a) Ts; (b ) pt; (c) hv ; ( d ) hs ; (e) tv; (/) uv. 

4. If an equilibrium mixture of water and steam is represented by point h of Fig. 
7:1, what is the quality of the mixture? See Art. 7:5 for data. Show approximate 
location of this state on Fig. 7:4. 

5. A triple-point mixture of ice, water, and steam has an enthalpy of 0 Btu and 
volume of 200 ft 3 per pound of the mixture. Determine its proportions. 

6. Based on the steam tables, tabulate the values of the pressure, enthalpy, vol¬ 
ume, entropy, and internal energy for a system consisting of 2 lb of saturated steam at 
the following temperatures: (a) 32°F; (6) 212°F; (c) 400°F; ( d ) 500°F; (e) 705.4°F. 

7. The same as Prob. 6, except that the system consists of 2 lb of saturated water. 

8. Tabulate the values of the temperature, enthalpy, volume, entropy, and internal 
energy for a system consisting of 0.5 lb of saturated steam at the following pressures: 
(a) 2 in. Hg abs; ( b ) 2 psig; (c) 14.7 psia; (d) 400 psia; (e) 500 psig; (/) 2000 psia; (g) 
3206.2 psia. 

9. The same as Prob. 8, except that the system consists of 0.5 lb of saturated water. 

10. For steam having a quality of 98 per cent, find the pressure (or temperature), 

the specific volume, the specific enthalpy, the specific entropy, and the specific internal 
energy when (a) t = 32°F; ( b ) p = 14.7 psia; ( c) p = 400 psia; ( d) t = 400°F. 


146 BASIC ENGINEERING THERMODYNAMICS 

11. An equilibrium mixture of steam and water has a specific enthalpy of 1000 Btu. 
Calculate its quality and its internal energy if (a) p = 2 psia; (b) t = 212°F; (c) t = 
400°F; (d) p = 3000 psia. 

12. A mixture of saturated steam and saturated water has a specific entropy of 
1.25. Calculate its quality if (a) p = 2 in. Hg abs; ( 6 ) t — 212°F; (c) p = 400 psia; 
(d) t = 650°F. 

13. A pound of wet steam has a specific internal energy of 900 Btu. What is its 
quality if (a) p — 2 in. Hg abs; ( b) p = 14.7 psia; (c) t = 300°F; ( d ) t = 500°F; (e) 
p = 3000 psia? 

14. Three pounds of steam is confined in a tank having an internal volume of 12 ft 3 . 
What is its quality or superheat if (a) p = 2 psia; ( b ) p = 14.7 psia; (c) t = 300°F; 
(d) t = 400°F; (e) p = 200 psia? 

15. Steam at the critical temperature and pressure is contained within a closed, 
rigid, and transparent tube. As the steam is cooled, its pressure and temperature 
decrease. Into what proportions does the meniscus divide the total internal volume 
of the tube as the following pressures and temperatures are reached? (a) p = 3000 
psia; ( 6 ) t = 600°F; (c) p = 700 psia; (d) t = 400°F; (e) p = 67 psia; (/) t = 200°F; 
(g) p = 2 in. Hg abs. 

16. What is the value of the psi function (</>) for saturated steam and for saturated 
water, as based on steam-table values, when t = 400°F? When t = 100°F? Write 
a general expression for \pf g . If the reference state at which the entropy and the 
enthalpy are zero were changed, would this expression still hold? For a given tem¬ 
perature or pressure would it give the same value for 

17. Calculate, based on steam-table values, the value of the zeta function (f) for 
saturated steam and for saturated water when t = 400°F; when t = 100°F. At which 
temperature does f have the higher value? What is always the value of f/ ff ? 

18. When water changes into steam at constant temperature, which is larger, the 
change of enthalpy or the change of internal energy? When ice I changes into water, 
which is the larger? 

19. Tabulate the values of the volume, the enthalpy, the entropy, and the internal 
energy of a 1 -lb superheated-steam system when the pressures and temperatures are 
as follows: (a) 1 psia and 120°F; ( 6 ) 98.2 psia and 400°F; (c) 98.2 psia and 408°F. 

20. For the following pairs of properties, determine whether the steam is saturated 
or superheated and its quality or superheat. If superheated, state its pressure and 
temperature, (a) p = 100 psia, h = 1200 Btu; (6) p = 400 psia, h = 1200 Btu; 
(c) t = 200°F, s = 1.8000; (d) t = 200°F, s = 1.7500; (e) v = 3 ft 3 , p = 100 psia; 
(/) v = 2 ft 3 , p = 500 psia; (g) h = 1300 Btu, s = 1.5000. 

21. What is the mean specific heat at constant pressure for superheated steam at 
the following states? (a) 2 psia and 200°F; (5) 2 psia and 500°F; (c) 14.7 psia and 
500°F; (d) 200 psia and 500°F; (e) 200 psia and 700°F; (/) 500 psia and 700°F. 

22 . What is the value of c Vi at the states listed in Prob. 21? 

23. What is the instantaneous value of the specific heat at constant pressure for 
steam of quality x? 

24. For steam at 300°F, x = 0.80, calculate the instantaneous value of the specific 
heat at constant volume. Use an interval of (a) 4°F; ( b ) 20°F. If your answers do 
not agree, give an explanation of the reasons for the discrepancy. Which answer do 
you consider to be the more accurate? Why? 

25. For the following paired pressures and temperatures, compute the volume, 
enthalpy, and entropy of 1 lb of compressed water. Use the unabridged steam tables. 
(a) 800 psia and 32°F; ( b ) 4000 psia and 32°F; (c) 800 psia and 400°F; ( d ) 4000 psia 
and 400°F; ( e ) 4000 psia and 700°F; (/) 1900 psia and 260°F. 


THE PURE SUBSTANCE 


147 


26. \\ ork Prob. 25 on the assumption that water is incompressible, basing your 
calculations on data taken from Table 1 of the steam tables. 

27. Saturated water at 400°F enters a pump which compresses it continuously to 

2000 psia. Determine the work per pound pumped if the compression is isentropic 
(a) by calculation based on Table 4 of the steam tables, ( b ) by the use of Fig. 3 of the 
steam tables, and (c) from dh = v dP. What is the temperature of the water as it 
leaves the pump? * 

28. A system consists of an equilibrium mixture of 3 lb of ice, 2 lb of water, and 
1 lb of steam. What are its temperature, pressure, volume, enthalpy, entropy, and 
internal energy? 

29. A system consists of three-fourths ice, one-fourth steam at a uniform tempera¬ 
ture of 12°F. What is its pressure? What, per pound of the mixture, are its 
enthalpy, volume, entropy, and internal energy? 

30. Based on Fig. 7:5, show that the value of the zeta function for saturated water 
or steam, or superheated steam, as based on steam-table values of the properties, 
must always be negative. Is £* for the state denoted by point y (on the saturated- 
solid line of Fig. 7:5) positive or negative? Why? 

31. Explain how Fig. 9 of the unabridged steam tables can be put to good use in 
speeding the solution of Prob. 20 above. 

32. Steam enters a turbine at 400 psia, t = 600°F, and expands adiabatically, leav¬ 
ing at a final pressure of 2 psia. The differences of elevation and of velocity between 
entrance and exit are negligible, (a) What maximum amount of shaft work may be 
performed per pound of steam flowing, and what is the final condition (quality or 
superheat) of the steam? ( b ) If the shaft work is 20 per cent less than the maximum, 
what is the final condition of the steam? Solve this problem both by calculation 
based on data taken from Tables 2 and 3 of the steam tables and by making use of the 
Mollier diagram for steam. Compare the two answers and the ease with which they 
were obtained. 

33. The same as Prob. 32, except that the steam enters at a velocity of 100 fps and 
leaves at 800 fps. 

34. A throttling calorimeter is used to determine the quality of the steam in a line 
in which the steam pressure is 300 psig. If the pressure and temperature in the 
expansion chamber of the calorimeter are as listed below, is it possible to fix the quality 
in the line? (a) 14.7 psia and 212°F; ( b ) 14.7 psia and 240°F; (c) 5 psia and 180°F. 
Where it is possible, determine the quality of the steam in the line, obtaining your 
answer both by calculation based on Tables 2 and 3 of the steam tables and by making 
use of the Mollier diagram for steam. Compare the answers and the ease with which 
they were obtained. 

35. A closed system consisting of a pound of steam at 67 psia, quality of 0.50, 
expands isothermally and reversibly to a final pressure of 20 psia. What amounts of 
work and of heat flow accompany the process? Show the approximate state path on a 
pv diagram and on a Ts diagram for steam. How are the work and the heat flow 
shown on these diagrams? 

36. The system of Prob. 35 undergoes a constant-volume reversible process as a 
result of which its pressure becomes 20 psia. What are its final temperature and 
quality (or superheat)? What amounts of work and heat flow accompany the 
process? Show the approximate state path on pv and Ts diagrams for steam. 

37. The system of Prob. 35 expands reversibly and at constant pressure until its 
volume is tripled. What is its final temperature? What work and what heat flow 
accompany the process? Show the approximate state path on pv and Ts diagrams for 
steam. 


148 


BASIC ENGINEERING THERMODYNAMICS 


38. The system of Prob. 35 is compressed reversibly and adiabatically until its 
pressure becomes 200 psia. Find the final temperature and quality (or superheat). 
What work accompanies the process? Show the approximate state path on pv and Ts 
diagrams for steam. 

39. A Carnot engine contains 1 lb of H 2 0. At the beginning of the isothermal 
expansion, it is a saturated liquid at 200 psia and, at the end, saturated steam (x = 1). 
The adiabatic expansion is carried to a pressure of 1 psia. What is the efficiency of 
the cycle? What net work is performed per cycle? What is the state (quality or 
superheat) of the working substance at the end of the adiabatic expansion and at the 
beginning of the adiabatic compression? Plot this cycle on a Ts diagram for steam. 

40. In steady flow around an open-system circuit, a pound of H 2 0 passes through 
the following four processes before completing the cycle: (1) Saturated water at 1 psia 
enters, and compressed water at 200 psia leaves, a liquid pump; the process is isen- 
tropic. (2) The water, as it leaves the pump, enters a heat exchanger and is heated 
reversibly at constant pressure until it becomes saturated steam. (3) Leaving the 
heat exchanger, the steam expands adiabatically and reversibly in an engine to a 
pressure of 1 psia. (4) A second heat exchanger closes the cycle as it condenses the 
steam reversibly and at constant pressure. What amounts of work and of heat flow 
accompany each of the four processes? What is the net work of the cycle? What 
is the efficiency of the cycle? Plot the cycle on a Ts diagram for steam. Compare the 
cycle and its efficiency with the results of Prob. 39. 

41. A pound of ammonia (see Appendix for tables of the properties of ammonia) 
passes through a cycle composed of the following four open-system steady-flow proc¬ 
esses: (1) Originally a saturated liquid at 180 psia, it is throttled adiabatically through 
a small opening into a space where the pressure is 30 psia. (2) It passes through a 
heat exchanger, receiving heat reversibly and at constant pressure until it is a satu¬ 
rated vapor. (3) The vapor enters a compressor and is compressed isentropically to 
a pressure of 180 psia. (4) The vapor leaving the compressor passes through a heat 
exchanger in which it is condensed reversibly and at constant pressure. What 
amounts of heat and work accompany each of the processes? What is the net work 
of the cycle? What amount of heat does the pound of ammonia receive in the course 
of the cycle? At what temperature does it receive this heat? What amount of heat 
does it reject? What is the lowest temperature at which heat is rejected? On a 
Ts diagram which shows the liquid and saturated-vapor lines for ammonia, show the 
appearance of this cycle. Is it a reversible cycle? 

42. A pound of saturated mercury vapor at'a pressure of 160 psia expands adiabati- 
cally and reversibly to a pressure of 2 psia. If the expansion takes place as a closed- 
system (nonflow) process, what work is performed? What is the final condition 
(quality or superheat) of the mercury? Answer the same questions when the 
process is of a steady-flow character through an open system. 

Symbols 

c p specific heat at constant pressure 
c v specific heat at constant volume 
h enthalpy of unit mass 
J proportionality factor 
p pressure, psi 
P pressure, psf 
s entropy of unit mass 
t scalar temperature 
T absolute temperature 


THE PURE SUBSTANCE 


149 


u internal energy of unit mass 
v volume of unit mass 
W work 

x mass of vapor per unit mass of mixture 
Subscripts 

f of the saturated liquid 
g of the saturated vapor 
fg change during vaporization 

i of the saturated solid; also, instantaneous value of the specific heat 
ig change during sublimation 
m mean value of the specific heat 
p constant pressure 
sat saturation 

v constant volume 


CHAPTER 8 


GENERAL THERMODYNAMIC EQUATIONS 
FOR THE PURE SUBSTANCE 


8:1. The Maxwell Relations. It will be noted that the primary prop¬ 
erties ( p , v, and T) are of a character such that it is possible to measure 
their values, corresponding to a given state of the pure substance, in a 
direct and tangible manner. The secondary properties, such as internal 
energy, enthalpy, and entropy, cannot, on the other hand, be directly 
measured, and it becomes desirable to establish relations that will make 
it possible to express their values in terms of the primary properties. 
Assume that, as the result of experiment and research, an equation has 
been developed that connects the pressure, volume, and temperature of a 
pure substance. 1 It will be remembered that this equation defines a pvT 
surface that will include all equilibrium states within the scope of the 
equation. The location of any point on this surface identifies an equilib¬ 
rium state of the substance and, with it, the values of all of the properties, 
primary or secondary. 2 As we move from one point on this surface to a 
second point, also on the surface, the change of any one property will be 
the same without regard to the path we may choose to follow between the 
two points. This is the foundation that underlies the use of Eqs. (1:4) 
and (1:5) in the identification of a property. Conversely, if we know P , 


x, and y to be properties, we are justified in writing 


dM dN 


[see Eq. 


dy dx 

(1:5)] as a true and correct relation between those properties. 

Suppose we select the equation that represents the application of the 
First Law to a simple closed system of unit weight as an example. This is 
[see Eq. (2:5)], J dQ — dW = J du. dQ and <9TT are not properties but 
can be expressed in terms of properties if we assume a maximum-work 
process; the corresponding equivalents are T ds and P dv. Substituting 
these values and rearranging, 


du = T ds 


Pdv 

J 


1 This equation, here called the primary equation of state, will be adequate for 
separate phases (though a separate equation may be required for-each phase) but will 
not normally apply to mixtures of phases since the pressure and temperature are not 
independent properties of such mixtures. 

2 The values of the secondary properties may be relative to some datum level, or 
levels, which is, as a matter of course, held constant. 

150 





GENERAL THERMODYNAMIC EQUATIONS 


151 


Comparing, we see that this equation is in the form of Eq. (1:4) with 
M = T and N = —P/J. The equation is expressed entirely in terms of 
properties (with the exception of the constant, J) and, following Eq. (1:5), 
we may write 


dM _ dN 
dy dx 



(8:1) 


Returning to the equation from which Eq. (8:1) was derived, we observe 
that, if the volume is constant, dv is zero and the second term on the right 


side of that equation becomes zero. 



Then {du) v = (T ds) v , or 
= T (8:2) 


Similarly, when the entropy is constant, ds = 0, and 



(8:3) 


(For an illustration of this relation, see Fig. 2:8.) 
manner, 


h = u + 


Pv 

T 


dh = du + 


P dv 

~T 


Proceeding in a similar 

vdP 
^ J 


But 


and, substituting, 
from which we obtain 


T ds = du + 

dh = T ds T 


P dv 

v dP 

~T 



(8:4) 


and, holding s and P constant, successively, 


and 



(8:5) 


( 8 : 6 ) 


Applying the same methods to the psi property, 
\p = u — Ts 

/ , Pdv 

d\p = du — T ds — s dT = du — I du H —-j 


sdT = 


P dv 


- sdT 
















152 


BASIC ENGINEERING THERMODYNAMICS 


and 

1 ( dP \ = ( ds \ 
J\dTj v ~ \dvj 

M). - - 

dA = _ P 
. dv ) T J 

From the definition of the zeta property as 

Pv 


r 


I = U + 


J 


Ts 


dr = du+ ^ - T ds - s dT = ( du + 


J 


J 


P dv 

AT 



(8:7) 

(8:8) 

(8:9) 



- sdT 


( 8 : 10 ) 

( 8 : 11 ) 

( 8 : 12 ) 


Equations (8:1) to (8:12) are known as the Maxwell relations. They 
furnish a basis from which many other equations of similar nature may be 
developed. For example, comparing Eqs. (8:2) and (8:6), we see that 


/diA _ / dh\ 
\ds) v \ds/p 


(8:13) 


and, making other similar comparisons, 



(8:14) 

(8:15) 

(8:16) 


Example 8:1. Check, approximately, Maxwell relations (8:1) to (8:12) for super¬ 
heated steam at 90 psia and 380°F. 

Solution. Table 3 of the Keenan and Keyes tables will be used. It will be noted 
that it will be desirable to extend the tabulation to cover the values of the internal 

















GENERAL THERMODYNAMIC EQUATIONS 153 


energy, the psi, and the zeta properties. At the designated state the calculation of 
these properties is as follows: 


u = h — 


Pv 

J 


1218.6 


(144) (90) (5.359) 
778 


1218.6 - 89.3 = 1129.3 Btu 


P = u - Ts = 1129.3 - (840)(1.6523) = 1129.3 - 1388.0 = -258.7 Btu 
r = h - Ts = 1218.6 - 1388.0 = -169.4 Btu 


To speed our calculation it will be convenient to extend the tabulation to include 
surrounding states. The result is shown below: 


V 


370°F 

380°F 

390° F 

89 psia 

V 

5.346 

5.422 

5.497 


h 

1213.4 

1218.7 

1224.0 


s 

1.6473 

1.6537 

1.6600 


u 

1125.4 

1129.4 

1133.4 


p 

-241.9 

-259.7 

-277.6 


r 

-153.9 

-170.4 

-187.0 

90 psia 

V 

5.284 

5.359 

5.434 


h 

1213.2 

1218.6 

1223.9 


s 

1.6459 

1.6523 

1.6586 


u 

1125.2 

1129.3 

1133.3 


+ 

-240.9 

-258.7 

-276.5 


r 

-152.9 

-169.4 

-185.9 

91 psia 

V 

5.223 

5.298 

5.372 


h 

1213.0 

1218.4 

1223.7 


s 

1.6445 

1.6509 

1.6572 


u 

1125.1 

1129.2 

1133.2 


p 

-239.8 

-257.6 

-275.4 


r 

-151.9 

-168.4 

-184.9 


(a) Checking Eq. (8:1) 

/0T\ = __ 1 /dP\ 

\dV ) s J \ ds ) v 

Here s and v are successively held constant. The value of s at the designated state 
is 1.6523. At 89 psia, this value falls between 370 and 380°F, interpolation showing 
the temperature to be approximately 377.8°F with a corresponding volume of 5.405 ft 3 . 
At 91 psia, interpolation shows the temperature and volume corresponding to this 
entropy to be, respectively, 382.2°F and 5.314 ft 3 . Then 

(dT\ _ 382.2 - 377.8 _ 4.4 _ 

\dy ) s 5.314 - 5.405 -0.091 

Similarly, the value of v at the designated state is 5.359 ft 3 , and this corresponds, at 
89 psia, to a temperature of 371.7°F and entropy of 1.6484 and, at 91 psia, to a tem¬ 
perature of 388.3°F and entropy of 1.6561. Then 

1 fdP\ -144(91 - 89) _ 

J\dsj v 778(1.6561 - 1.6484) 



















154 


BASIC ENGINEERING THERMODYNAMICS 


( dT\ 

— J above, the relatively small discrepancy may be 

attributed to the small number of significant places in the differences that enter into 
these calculations. 

(b) Eq. (8:2): = T. Here v is again constant as in the second half of part a. 

\ds / v 

At 89 psia, 371.7°F (see part a), the internal energy is 1126.1 Btu, and, at 91 psia, 
388.3°F, u = 1132.5 Btu. The corresponding entropies may be taken from part a. 


Then 


du\ 1132.5 - 1126.1 001 m oon , CylA 

— ) = 1 c aoa = 831; T = 380 + 460 = 840 

ds J v 1.6561 — 1.6484 


( 

( (97/\ P 

—J = — -j- Here s is constant. At 89 psia, 377. 8 °F, and at 

91 psia, 382.2°F (see part a), the internal energies are, respectively, 1128.5 and 
1130.0 Btu. The corresponding volumes may be taken from part a. Then 


dv ) 


, = 1130 A - 1128 ; 5 = -16.5; - % = = -16.7 


5.314 - 5.405 


( 

(d) Eq. (8:4): (^)_ = j (^) p - From part a, 

(iT\ 382,2 - 377.8 

144(91 - 89) 00153 


778 


Directly from the table above, 


1 ( dv \ _ ( 1 \ ( 5.434 - 5.284 \ 
J \ds)p \77Sj \ 1.6586 - 1.6459/ 


= 0.0152 


(e) Eq. (8:5): At 89 psia, 377. 8 °F, the enthalpy is 1217.5 Btu, and, at 

91 psia, 382.2°F, h = 1219.6 Btu. Then 

5.359 


/ dh\ _ 1219.6 - 1217.5 v _ 5.35S 

\dPj. 144(91 - 89) 7d; J 778 

(/) Eq. ( 8 : 6 ): = T. Directly from the table, 

O, = B -^- 1 '645 9 = 842 ' ^ = 840 

WEq. (8:7):i(^ = (| 

1 fdP\ = 144(91 - 89) 

J\dTj v ~ 


= 0.0069 


) 


From data computed in part a, 


= 0.0223 


778(388.3 - 371.7) 

Directly from the table, 

(£) ~ ~ = 0.0226 
\dvjr 5.298 — 5.422 

( h ) Eq. ( 8 : 8 ): = —s - At 89 psia, 371.7°F, \p is —244.9 Btu, and, at 91 psia, 

388.3°F, f = -272.4"Btu. Then 


= -272,4 - (-244.9) _ 

\dTj v 388.3 - 371.7 


■s = -1.6523 













GENERAL THERMODYNAMIC EQUATIONS 


155 


O Eq. (8:9): T = “ Directly from the table, 

(<N\ = -257.6 - (-259.7) 

\dv ) t 


5.298 - 5.422 


= -16.9; - j = -16.7 


O’) Eq. (8:10): j (jpp'j p ~ ~ (jlp) ' ^ data may be taken directly from the 

table. 


1 ( dv\ ( 1 \ Z5.434 - 5.284\ 

j{sT) P " \ 778 / (~ 39 0 ~ — 370 ) = 0 0000096 = 


fds\ _ 1.6509 - 1.6537 _ _ AAAAn _ 

VdP/r 144(91 - 89) 0.0000097 

( k ) Eq. (8:11): (i), = — s. Substituting values read directly from the table, 

(K\ = -185.9 - (-152,9) _ _ , ft , 9 o 

\dTjp 390 - 370 Lb5, 1.6523 

(l) Eq. (8:12): = p Again directly from the table, 

(dt\ _ -168.4 - ( -170.4) A AAgA , i> A AAAA 
\dp) T 144(91 - 89) 0.0069, j 0.0069 

8:2. The specific heats at constant volume and constant pressure are 
themselves derivatives of properties. The specific heat at constant vol¬ 
ume has been defined (see Chap. 2) as c v = (jpp'j * Based on Eq. (8:2), 


( 


or 


du\ _ / du\ / ds\ _ rp ( ds\ 

df) v ~ \ds)v\df) v ~ WA 


Cv l /a:r) v T \dTJ v 


(8:17) 


Similarly, employing Eq. (8:6), 

C » *= (S') 


= T 


ds 

Fr> 


(8:18) 


Equations (8:17) and (8:18) express the specific heats in terms of the 
entropy. The change of the specific heats with respect to changes in the 
primary properties may also be derived. From Eqs. (8:17) and (8:7), 




and 


ds\ 

dv) T J \dT/ v 


Then, 












156 


BASIC ENGINEERING THERMODYNAMICS 


Proceeding in a similar manner, but making use of Eqs. (8:18) and (8:10), 
it may be shown that 



( 8 : 20 ) 


By means of Eqs. (8:19) and (8:20) and a primary equation of state, a 
limited number of specific heat data may be extended to cover a wider 
range of states. 

We learn from the calculus that the relation between the partial deriva¬ 
tives of any three point functions 1 is 



( 8 : 21 ) 


This equation may be used to good effect to establish certain interesting 
and useful relations between properties. For example, the ratio of the 
specific heat at constant pressure to the specific heat at constant volume, 
c p /c v (usually called simply the ratio of the specific heats and designated by 
the symbol k), may be expressed in terms of derivatives of the properties 
with its aid. If we select the primary properties as the point functions, 
Eq. (8:21) becomes 


From Eq. (8:17), 


Then 



( 8 : 22 ) 


/ dP\ _ C* / dT\ (dP\ = c v (dP\ 
\dT/ v T\dsJ v \dTj v T\ds) v 

But, from Eq. (8:1), 



and, substituting above, 


Similarly, 


(dP\ _ Jc v (d_T\ 
\dT/ v T \dv) s 


c p = T 


(8:23) 


(*l) 

V'tJ, 


or 


T 

c 


; (w), -' 


(dT\ = T / ds\ /dT\ = T 
\dv)p Cp\dT/p\dv)p c p \dv) } 


1 A point function is a coordinate of a point. When three point functions are 
involved, the point lies on a surface such as the pvT surface. Properties are therefore 
point functions, and, for the pure substance, three are necessary to establish the 
location of the points. 















GENERAL THERMODYNAMIC EQUATIONS 


157 


and, making the substitution suggested by Eq. (8:4), 

(dT\ = JT_ / dP\ 

\dv)p Jc p \dT/ s 


(8:24) 


Making the substitutions in Eq. (8:22) that are suggested by Eqs. (8:23) 
and (8:24), we have 

_ Jc v (dT\ T (dP\ (dv\ 

T \dv) s Jc p \dT) s \dP) T 


and, solving for c p /c v , we obtain 

, = Cp = / dT\ tdP\ / dt/ \ = / dP\ ( dv\ 
C v \dv / s \dTJ s \dP /t \dv/ s \dP/T 


(8:25) 


Example 8:24. Estimate, for superheated steam at 90 psia, 380°F, the specific 
heats at constant volume and at constant pressure and the ratio of the specific heats, 
using Eqs. (8:17), (8:18), and (8:25). Compare with the corresponding values as 
obtained directly from the definitions of these properties. 

Solution: 


(a) Eq. 


(8:17): 



Using data from Example 8:1, the volume is the 


same at 89 psia, 371.7°F, and at 91 psia, 388.3°F, as at the designated state. At the 
first of these two states the entropy is 1.6484 and, at the second, 1.6561. Then 


„„„ ( 1-6561 - 1.6484^ n on 
= 840 ( 388.3— «u) = °- 39 

But c. = (^) l> y definition, or, again using values from Example 8:1, 


1132.5 - 1126.1 
C * = 388.3 - 371.7 = °' 386 


(6) Eq. (8:18): c v = T (jf) , or, using the table of Example 8:1, 

/ 1.6586 - 1.6459\ n coo 
C - = 84 H 390 — 370 / == °- 532 

By definition, 

dh\ = 1223.9 - 1213.2 


dT/ j 


390 - 370 


= 0.535 


c p — 

(c) Eq. (8:19): k = (Jp) T - A S ain usin S the data of Exam P le 8: b 


k = 


144(91 - 89) 1 T 5.298 - 5.4221 _ 
5.314 - 5.405 J L 144(91 — 89) J 


Based on the values of c p and c v as computed from Eqs. (8:18) and (8:17), 



0.532 


1.36 


0.39 














158 


BASIC ENGINEERING THERMODYNAMICS 


Based on the definitions of c v and c v , 
0.535 


k = 


0.386 


= 1.38 


Example 8:2 B. Show that, for a gas having a primary equation of state in the 
form of Eq. (1:6), the internal energy, the enthalpy, the specific heat at constant 
volume, the specific heat at constant pressure are all functions of the temperature 
alone, i.e., are independent of changes in the pressure or the volume. 

Solution: 

jP civ 

(a) For the reversible process, du = T ds - -j-" Dividing through by dv and 

writing the equation as it applies to a reversible isothermal, 


(-) - 

\ dv ) T 


( 


ds\ _ P 
dv ) T J 


But, from Eq. 8:7, 

/ds\ = 1 (dP 
\dv) t J \dT 

and, substituting, 


). 


( 


^ - 2 (g). - * 


dv ) 


J 


J 


The equation of state is pv = CT or P = 1 44CT/v, and 

144 C 


(du\ = T/dP\ _ P _ T 144 
\ dv ) T J \dT / v J J V 


p p p 
J = J - J = 0 


The internal energy of this gas is therefore independent of volume changes if T is 

du\ 


constant. 


A1S0 > (l0r = (^)r(^X =OX (^)r = °’ “ iS ak ° inde ' 

pendent of pressure changes at constant T. 1 

( 6 ) It has been shown in part a that u is a function of temperature alone. But 
h = u + Pv/J, and Pv/J( = 144pv/J) is also a function of the temperature alone, the 
product pv having a constant value for this gas if the temperature does not change. 
Therefore the enthalpy is independent of pressure and volume changes except as a 
change in the product pv indicates that a change in temperature has taken place. 

(c) From Eq. (8:19), 


/aoq _ T(?P\ = T 
\ dv J t J\9T‘J, J ' 


and 


/ dc v \ = I = 0 

\dP/ t \dV/T\dP/T 


(d) Similar reasoning, based on Eq. (8:20), will show that c p is, for this gas, independ¬ 
ent of pressure and volume. 


( 


1 AS the preSSUre a PP™ aches (B) T infinity. At 

—5 ) therefore is not necessarily zero. 
dr / t 


zero pressure. 





GENERAL THERMODYNAMIC EQUATIONS 


159 


Another example of the application of Eq. (8:21) to thermodynamic 
purposes is found when it is written in the form 

fdT\ 

) P * s rec iP roca ^ the specific heat at constant pressure, or l/c p . 

The second partial derivative is the reciprocal of a ratio which is readily 
obtained experimentally and which is known as the Joule-Thomson coeffi¬ 
cient; the symbol used to represent this coefficient is /ij. The experi¬ 
mental method employed is to allow the slow flow of a fluid through a 
horizontal tube which is fitted with a porous plug. The pressure and 
temperature upstream from the plug are held constant as the downstream 
pressure is gradually reduced. Provision is made for measuring pressure 
and temperature both upstream and downstream. As the downstream 
pressure is gradually reduced, the temperature below the plug that corre¬ 
sponds to each downstream pressure is noted and plotted as the ordinate 
with the pressure as the abscissa. The points so obtained are connected 
to form a curve. Application of the steady-flow energy equation [Eq. 
(3:5)], with the difference between upstream and downstream velocities 
assumed to be negligible, shows that for this adiabatic process (the tube 
is insulated to prevent appreciable heat flow) the upstream and down¬ 
stream enthalpies are equal. The slope of the curve may therefore be 

expressed as ( ~p ) and gives the value of hj, the Joule-Thomson coeffi¬ 
cient. The Joule-Thomson coefficient is usually positive (the change of 
temperature across the plug usually being negative), but exceptions are 
found. 

The Joule-Thomson apparatus may be adapted to the experimental 
determination of another property derivative by including a heating (or 
cooling) element in the porous plug. By adjusting the rate of heat flow, 
the downstream temperature is held at the upstream value so that the 
process now takes place at constant temperature. The flow is no longer 
adiabatic, and Eq. (3:5) indicates that the heat flow per pound of fluid 
flow is numerically equal to the change of specific enthalpy across the 
plug. If this heat flow is measured by some suitable means and plotted 
as the ordinate with the pressure as the abscissa, the slope of the resulting 

curve is • This slope is called the constant-temperature coefficient; 

we shall denote it by the letter c. For the same substance at the same 
state, c will always have a sign opposite to that of hj. 

Substituting the equivalent symbols in Eq. (8:26), 

1 1 


Cp f-i j 


c = — 1 


or c p =- 

UJ 


(8:27) 







160 


BASIC ENGINEERING THERMODYNAMICS 


This equation may be used as a check on the values of c p as obtained by 
other means. 


Example 8:2 C. From steam-table values, estimate the Joule-Thomson and the 
constant-temperature coefficients for steam at 90 psia, 380°F. Compute the specific 
heat at constant pressure based on Eq. (8:27), and compare with the values as obtained 
in Example 8:2 A. 

Solution: 

(a) At 90 psia, 380°F, h = 1218.6 Btu. At 80 psia, this enthalpy corresponds to 
a temperature of 376.8°F and, at 100 psia, to 383.3°F. The Joule-Thomson coeffi¬ 
cient may be calculated from these values as 


hj 


SdT\ 376.8 - 383.3 
\dP) h ~ 80 - 100 


0.325°F/psi 


(6) A similar calculation of the constant-temperature coefficient, based on the same 
pressure range, is 


fdh\ 1220.3 - 1216.8 

\dPj T ~ 80 - 100 


—0.175 Btu/psi 


(c) From Eq. (8:27), 



(-0.175) 

0.325 


0.538 


The comparative values from Example 8:2A are 0.532 and 0.535. 


8:3. The General Equations for Enthalpy, Internal Energy, and 
Entropy. We have applied the adjective primary to the properties that 
can alone be directly and tangibly measured and to the equation of state 
that connects those properties. This equation of state is customarily 
expressed either in the form v = f(p,T) or as p = f(v f T). If the primary 
equation of state is as simple as Eq. (1:6), either of these two forms may, 
of course, readily be converted to the other but that is not generally the 
case. As an example, consider the primary equation of state for steam 
that is presented on page 15 of the Keenan and Keyes tables, and con¬ 
template the difficulty of changing it to the form p = f(v,T). 

We now face the problem of developing secondary equations of state to 
apply to the intangible properties of enthalpy, internal energy, entropy, 
etc., expressing these properties in terms of the primary properties and 
their derivatives. The Maxwell relations furnish a set of tools to aid in 
the accomplishment of this purpose. The particular tool or tools which 
we select for the purpose will depend upon which form of the primary 
equation of state is available to us and also upon the character of addi¬ 
tional information (such as experimentally determined values of the 
specific heat at constant pressure or of the Joule-Thomson coefficient) 
that is at hand. 

For example, we may set up a general equation of state expressing the 
enthalpy as shown below. It is a principle of the calculus that dy = 





GENERAL THERMODYNAMIC EQUATIONS 161 


(8), dx + (2), 


dz when three variables are involved. Then 


,, P dv v dP m , . v dP 

dh = du -\ - r -1-— = T ds + 


J 


J 


J 


= c p dT — —T 


(ft) 


- t (ft), 


dT + T 


(s) 


dP + l ~ 

T J 


V 

T J 


dP 


Substituting 


1 / dtA 
J \dT/j 


for 


and, integrating, 


dh = c p dT — T 


t 1 rp 


(| p) T ^ see Eq ’ ( 8:10 )1> 

/ dv\ ] dP 

\df)p 


(8:28) 


h = I CpdT — ^ 

T i J J Pi 


r (ft), - ■ 


dP *T hi 


(8:29) 


in which hi is a constant of integration, being the enthalpy at some refer¬ 
ence level at which the pressure is Pi and the temperature T\. A study 
of the procedure used in the development above will indicate that the 
total change of enthalpy has resulted from two steps, the first term of Eq. 
(8:29) giving the change of enthalpy that takes place as the pressure is 
held constant while the temperature changes to its final value and the 
second term the increment of enthalpy as the temperature remains con¬ 
stant while the pressure is adjusted to its final value. For the purpose of 
the integration indicated in the second term of that equation, T is there¬ 
fore handled as a constant. This equation would apply most conveniently 
when the primary equation of state was in the form v = f(p,T). 


Example 8:3d. Develop an expression for the change of enthalpy of a gas with a 
primary equation of state in the form pv = CT. 

Solution: 


h ’~ h ' = Ir>’ dT -7 Ir‘[ T (I?) F ~ v l dP 

For this gas, v = and T — v = T — v = v — v = 0. 

tion between pressure limits therefore vanishes and 


hi — h\ 



or if c p is constant, 


hi — hi = Cp(T i — T i) 


The integra- 


The general equation for internal energy may be obtained in a similar 


manner: 














162 


BASIC ENGINEERING THERMODYNAMICS 


du = Tds - - y - = T ( 


w). iT + T (S), 


dv — 


P dv 

~r 


Substituting 


= c v dT + 
1 / dP 


P 


T 'Si, j 


ds 


dv 


j vary. for (Si), [see Eq ' (8:7)] ’ 


du = c v dT -f- -j 




dv 


(8:30) 


and, integrating, 


u = 


c v dT H —j 

T i J 


v (dP 

rji i 


vi 


dT J v 


- P 


dv + 


(8:31) 




Example 8:35. A primary equation of state is in the form P = _ ^ 

where 5, a, and 6 are constants. 1 Write an expression for the change of internal 
energy of a substance that obeys this pvT relation. 

Solution: 

( dP\ _ R A rn ( dP\ p _ RT RT , a 
\dr) v V - 6 and T (dr), v-b v-b + v 2 

Then, from Eq. (8:30), 

u 2 - Ul = f Tl c v dT +- f V2 — dv = f T ' c v dT 

JTi J Jv 1 y 2 Jt 1 J \y 2 yi/ 

Proceeding to the development of a general equation for entropy, we 

observe that, from Eq. (6:8), ds = — + -fff' Substituting the value of 

1 1 

du as expressed in Eq. (8:30), 


_ du P dv _ c v dT 1 

ds - y + ~jt Y~ + J 


dP\ 

jt) v 


p 

T 


dv + 


P dv 

TT 


-C M lfdP} 

v v rp I T 


Integrating, 


J \dT) v 


dv 


dT l 
Tl ° v T + J 



dP\ 

dT) v 


dv + Si 


(8:32) 


(8:33) 


Equations (8:30) to (8:33) are examples of forms of the general equation 
most conveniently used when the primary equation of state is in the 
form p = f(v,T)‘ Note that the general method of development has 

1 This is the van der Waals equation. It represents an attempt to fit a single form 
of primary equation of state to a wide range of pure substances, the constants being 
based on the temperature, pressure, and volume at the critical state of the substance. 
The equation will be discussed in greater detail in a later chapter. 



























GENERAL THERMODYNAMIC EQUATIONS 


163 


differed from that followed in developing Eqs. (8:28) and (8:29) only in 
the use of a constant-volume step, instead of a constant-pressure, for 
the measurement of the first increment of the property. 

If we start with dh = T ds -j- v dP/J (see Art. 8:1), alternate expres¬ 
sions for ds and s may be obtained in which the increments of entropy 
are measured over constant-pressure and constant-temperature segments. 
These are 

, dT 1 / dv\ „ 

s - c p ~ j \sf) p dP (8:34) 

and 

s = II c »f~ 7 il (5), dP + “ ( 8:35 ) 


Example 8:3 C. Develop an expression for the change of entropy that is based on 
the van der Waals equation (see Example 8:3 B). 

Solution. The van der Waals equation is in the form p — f(v,T). Equation (8:32) 
will therefore be preferable to Eq. (8:34) in the solution of this example. 


s 2 — Si 


/, 


T* dT 
Tt C "-f + 


in 


if).*' 


f^ 2 dT 1 f Vi Rdv 

r, c * ~f + J J V1 


and if c v is constant, this becomes 


s 2 — Si = c v log 


T <i R . Vi 

e TFT + T l°g 


T i 1 J 


Vi — b 


8:4. The Clapeyron relation is based on the Maxwell relation that 
is stated as Eq. (8:7) and applies this relation to mixtures of phases for 
which the pressure and temperature are not independent properties, i.e., 
the pressure is a function of the temperature alone without regard to a 
variation of the other properties. This makes it possible to substitute 



becomes 


in Eq. (8:7) for such mixtures, and the Maxwell relation 


J 



dP 

dT 


When a change takes place at constant temperature (and, of course, 
constant pressure), 



S 2 — s i _ 1 dP 
v 2 — Vi J dT 


(8:36) 


where the subscripts refer, respectively, to the saturated states of the 
two phases. Thus, if the change is from a saturated liquid to a saturated 
vapor, s 2 — Si = s g — Sf = S/ g and v 2 —■ V\ = v g — Vf = v/ g . Substitut¬ 
ing in Eq. (8:36), 












164 


BASIC ENGINEERING THERMODYNAMICS 


The application of Eq. (8:36) is not limited to changes between the 
saturated-liquid and saturated-vapor states but may be applied to any 
change of phase, partial or complete, of the pure substance that takes 
place at constant temperature. 

= T. For a two-phase mixture, T will be 

constant if P is constant, and we may write = ^• Then 

/ dh\ _ / dh\ / ds\ _ /ds\ _ T dP 

\dv / t \ds Jt \dv/ t \dv/ t J dT 

Making the equivalent change in Eq. (8:36), we may write 


From Eq. (8 


*»■ (2 


J 12 — hi _ T dP 
t> 2 — Vi J dT 


(8:37) 


It is Eq. (8:37) that is known as the Clapeyron relation. Its possible uses 
in connection with two-phase mixtures are obvious. 

Example 8:4. Check the steam-table values of (a) the enthalpy of vaporization 
at 90 psia and ( b) the enthalpy of sublimation at 20°F, using steam-table data for 
the pressure-temperature relation and the change of volume that accompanies the 
change in phase and applying the Clapeyron relation. 

Solution: 


(a) dP /dT will be calculated as AP/AT over the range 88 to 92 psia, or 


144(92 - 88) 
321.83 - 318.68 


dP 
dT 

, , , T dP . 

h 2 — hi — hf g — -j [Vg 


183 [Table 1, Keenan and Keyes tables] 

- v f ) = (^f^) (183)(4.896 - 0.018) = 895 Btu 


The steam-table value is 894.7 Btu. 


( b) AP/AT will be calculated over the range 15 to 25°F in Table 5 of the Keenan and 
Keyes tables. 

dP _ 144(0.0640 - 0.0396) 
dT~ 25 - 15 U ^ 51 

T rip / 480 \ 

hi - hi = hi B (v g ~ Vi) = ( — ) (0.351) (5658 - 0.02) = 1225 Btu 


The steam-table value is 1219.9 Btu. 


Problems 

1. Check, approximately, Maxwell relations (8:1) to (8:12) for superheated steam 
at 200 psia and 600°F. 

2. Using the data of Table 3 of the steam tables, estimate the specific heats at 
constant volume and at constant pressure and the ratio k for superheated steam at 
200 psia, as based on Eqs. (8:17), (8:18), and (8:25). 

3. Show in detail the proof that c v for an ideal gas is a function of temperature 
alone if it varies at all (see Example 8:2 B, part d). 

4. Based on the Maxwell relations, prove that lines of constant pressure on a 
Mollier diagram (a) are straight lines as they cross the saturated region for any vapor 








GENERAL THERMODYNAMIC EQUATIONS 


165 


and (b) are concave upward in the superheat region, (c) Prove that the critical point 
must be a point of inflection on the saturated liquid—saturated vapor curve as plotted 
on a Mollier diagram. 

5. W hat are the coordinates of each of the following charts? (a) Lines of constant 
entropy have a slope proportional to the volume of the system. (6) Lines of constant 
volume have a slope proportional to the absolute temperature of the system, (c) 
Lines of constant pressure have a slope proportional to the absolute temperature of the 
system. ( d ) Lines of constant temperature have a slope proportional to the volume 
of the system. 

6 . W hat Maxwell relation shows that the increase of kinetic energy of the fluid 
per unit decrease of its pressure in adiabatic and reversible steady flow through a 
horizontal nozzle is proportional to its specific volume? 


7. Prove that c v — c v — -j by following the procedure outlined 

below: 

(a) Write the expression for ds as the sum of the partial derivatives of s with 
respect to T and v. 

(b) Write the expression for dv as the sum of the partial derivatives of v with respect 
to T and P. 

(c) Substitute dv from ( b) in (a). 

(d) Write the expression for ds as the sum of the partial derivatives of s with respect 
to T and P. 

( e) Equate the coefficients of dT in (c) and (d), and multiply both sides of the 
equality by T. Substitute c p for T(ds/dT) P [see Eq. (8:18)] and c v for T(ds/dT) v 
[see Eq. (8:17)] in this equation. 

(/) Make the substitution in (e) which is suggested by Eq. (8:7), and rearrange to 
the required form. 

8 . Using the expression for the difference between the specific heats which was 
developed in Prob. 7, show that this difference is, for a gas having a primary equation 
of state in the form of Eq. ( 1 : 6 ), equal to 144 C/J. 

9. Calculate from steam-table data the values of the Joule-Thomson coefficient 
and the constant-temperature coefficient for steam at 200 psia and 600°F. Show that 
the values obtained are consistent with the specific heat at constant pressure as 
calculated in Prob. 2. 

10. In Eq. (8:21), substitute the internal energy, the volume, and the absolute 
temperature as the point functions, and obtain a relation analogous to Eq. (8:27). 
Show that one of the components of this relation is c v , and explain how the values 
of the others might be established experimentally. Are these experiments performed 
on an open or a closed system? Is the Joule-Thomson coefficient obtained by experi¬ 
ment on an open or a closed system? 

11. Would you classify the Joule-Thomson coefficient as a property of the system? 
Could it be used as a coordinate in plotting the state of the system? 

12. In a Joule-Thomson experiment, the upstream pressure and temperature of a 
gas are 20 psia and 35°F, respectively. The downstream pressure is 15 psia, and the 
downstream temperature is 35.6°F. It is also observed that if the temperature is to 
be the same at the two points, 0.41 Btu of heat must be removed per pound of gas 
flow between the two sections. What is the Joule-Thomson coefficient? The con¬ 
stant-temperature coefficient? The specific heat at constant pressure? 

13. Show that the Joule-Thomson coefficient and the constant-temperature coeffi¬ 
cient of an ideal gas must both be zero. When these values are substituted in Eq. 
(8:27), what value is obtained for c p ? 

14. Develop expressions for the change of enthalpy, of internal energy, and of 



166 


BASIC ENGINEERING THERMODYNAMICS 


entropy for a substance for which the primary equation of state may be expressed as 
P(v — b) = CT, in which b and C are constants. 

15. Develop expressions for the changes of internal energy and of entropy of an 
ideal gas. 

16. Check the relative values of s/ g , v/ g , and h/ g as shown in the steam tables for 
saturated steam at 200 psia, using the Clapeyron relation. 

17. At 50°F the saturation pressure of a vapor is 89.19 psia, and the change in 
specific volume during vaporization is 3.27 ft 3 . At 52°F the saturation pressure is 
92.66 psia and, at 48°F, 85.82 psia. What are the changes of enthalpy and of entropy 
during vaporization at 50°F? 

18. For a certain substance, the change of specific volume during vaporization at 
70°F and 49.62 psia is 1.574 ft 3 , and 148.9 Btu of heat is required to accomplish the 
vaporization of 1 lb reversibly. Estimate the saturation pressure at a temperature 
of 72°F. 

19. Prove that if the melting temperature of a substance increases with the pressure 
applied on its surface, the specific volume of the resulting liquid is larger than that of 
the solid with which it is in temperature-pressure equilibrium. 

20. Note that in Fig. 7:2 the boundary line that separates ice I from ice III is 
nearly horizontal. What does this indicate with regard to the change of enthalpy 
which accompanies the change from ice I to ice III? Also note that, at the triple 
point at which a mixture of ice I, ice III, and liquid water is possible, the slope of the 
fusion locus changes from negative to positive. What conclusions can you draw as 
to the change of volume as ice I changes to ice III? 

Symbols 

a, b constants 

c constant-temperature coefficient 
c p specific heat at constant pressure 
c v specific heat at constant volume 
h enthalpy of unit mass 
J proportionality factor 
k ratio of the specific heats, c p /c v 
p pressure, psi 

P pressure, psf; pressure in general 
Q heat flow 
R a constant 
s entropy of unit mass 
T absolute temperature 
u internal energy of unit mass 
v volume of unit mass 
W work 

Greek Letters 

£ zeta property of unit mass 
Hj Joule-Thomson coefficient 
\p psi property of unit mass 

Subscripts 

h constant enthalpy 
p constant pressure 
s constant entropy 
T constant temperature 
v constant volume 


CHAPTER 9 


THE PERFECT GAS 

9:1. The Gas and the Vapor. In preceding pages the terms gas and 
vapor have been used interchangeably. Let us now adopt a distinction 
between these terms. Many of the substances dealt with by the engineer 
remain in the gas phase during the series of processes through which he 
puts them. An example is the air that enters the cylinder of an internal- 
combustion engine or an air compressor; at no time during the cycle 
carried out in the apparatus does the air condense to liquid form or, 
indeed, even closely approach that phase. On the other hand, some 
substances change back and forth between separate phases in the course 
of the engineering cycle; the water that is the working fluid in the steam 
power plant furnishes an example. To distinguish between the two 
terms as we shall hereafter apply them, gas will be used when, in the 
course of the proposed process or processes, the substance will remain in 
the form of a gas; the use of vapor will indicate that a change in phase is 
contemplated. Thus the same substance may be classified as a gas for 
one set of operations, as a vapor for a second series of processes. When, 
in the course of the contemplated operations, the temperature remains 
above the critical temperature and the pressure never rises above the 
critical pressure, the classification as a gas would unquestionably be 
justified according to the convention we have adopted. Even if the 
temperature falls below the critical temperature but remains above the 
saturation temperature equivalent to the coexisting pressure, according 
to the pressure-temperature relation for the substance, the designation 
as a gas would apply. It will be observed that the classification as gas 
or vapor is closely connected with the pressure and temperature at the 
critical point as compared with the lowest temperature and highest 
pressure to be experienced during the proposed state changes. 

Many substances, including, for example, hydrogen, oxygen, and 
nitrogen, were formerly called permanent gases because no usual series of 
engineering operations would bring about their liquefaction; yet all of 
these so-called “ permanent ” gases may be changed to the liquid phase 
if sufficiently large decreases of temperature and (possibly) increases of 
pressure are applied. 

Many of the cycles soon to be discussed are gas cycles, i.e., at no point 
in the cycle does the substance become a liquid. As gases, all substances 

167 


168 


BASIC ENGINEERING THERMODYNAMICS 


behave in a somewhat similar manner; this is particularly true at pressures 
that are low relative to their critical pressure. In order to avoid divert¬ 
ing attention from more essential features in the coming study of the gas 
cycle, it will be convenient to assume a hypothetical substance having a 
simple equation of state, one which can be easily handled mathematically. 
Yet this equation of state should relate the pressure, volume, and temper¬ 
ature in a manner such as to reflect, as accurately as is practicable, the 
actual relations between these properties for the real gases with which the 
engineer works. That, briefly, is the reasoning behind the concept of 
the ideal, or perfect , gas. 

9:2. The Perfect Gas. The work of Robert Boyle with air systems at 
moderate temperature resulted in the statement that has come to be 
known as Boyle’s law: As a gas system is compressed or expanded at a 
constant temperature, the product of its pressure and volume will remain 
constant. In mathematical terms, Boyle’s law may be stated as 


{PV) T = c or S- 1 = - ( T const) (9:1) 

P 2 Vi 


Boyle’s investigations were made in the seventeenth century. The 
later work of Charles and others lead to the modern version of Charles’ 
law: 

1. If the pressure is constant during the expansion or compression of a gas 
system, the volume will vary directly with the absolute temperature. 

2. If the volume is constant, the absolute pressure will vary directly with 
the absolute temperature. 

Or, stated mathematically, 

1. At constant pressure, 


Vl 

'C 2 

2. At constant volume, 

Pi 

P 2 



(9:2) 


(9:3) 


More recent investigation has shown that the laws of neither Boyle 
nor Charles are exact over a wide range of states. However, at pressures 
that are low relative to the critical pressure and temperatures which are 
high relative to the critical temperature, both reflect quite accurately 
the behavior of the gas. Indeed, as they apply to hydrogen or helium, 
these laws supply a method for the determination of temperatures on the 
thermodynamic scale. As the temperature is lowered, the product PV 
in the Boyle’s law equation (9:1) will approach zero as a limit. The 
absolute zero of temperature corresponding to this limit must be found 



THE PERFECT GAS 


169 


by extrapolation, of course, since absolute zero of temperature cannot be 
attained, nor would any substance remain a gas at that temperature. 
The instrument employed is called a constant-volume or a constant- 
pressure thermometer according to whether the volume or the pressure 
of the gas system is held constant as the temperature is changed between 
arbitrarily defined anchor points of temperature. 

Air is an important gas from the engineering standpoint. The error 
incurred in applying the Boyle and Charles relations to air at atmospheric 
pressure is about one-tenth of 1 per cent and, at 300 psia, approximately 
1 per cent. For more exact computations, corrections are necessary, 
especially at high pressures, but, for the immediate purpose of developing 
approximate equations to apply to real gases, the accuracy of these laws 
is sufficient to justify their use in framing a primary equation of state 
for the so-called perfect gas. Moreover, Eqs. (9:1) to (9:3) present very 
simple relations between the primary properties, leading to an uncompli¬ 
cated primary equation of state. 

The restrictions on the use of the Boyle and Charles relations, if they 
are to give accurate results, to states in the low-pressure or high-tempera¬ 
ture range lead to the concept of the perfect gas as having an extremely 
large volume for a given number of molecules, so large, in fact, that the 
attractive forces between molecules vanish and energy storage in the 
molecular system is solely due to molecular mass and velocity (the 
kinetic energy of the molecule). Moreover, because they are so widely 
separated, collisions between molecules would not occur. 

The rate of storage of energy in the system composed of what we call a 
perfect gas may be calculated mathematically on the basis of the kinetic 
theory of matter if the velocity of the molecule is assumed to be entirely 
translational in character. This would be the case for a molecule com¬ 
posed of only a single atom; this molecule could be said to have three 
degrees of freedom of motion, the X, Y, and Z directions of Fig. 9T. 1 

1 Molecules composed of more than a single atom could have rotational and vibra¬ 
tional kinetic energy as well as translational. For instance, a diatomic (two-atom) 
molecule could have not only the three degrees of freedom of translational motion but 
could also rotate about either of two axes perpendicular to each other and to a line 
joining the two atoms. This would be classified as two additional degrees of freedom 
of motion, or a total of five for the diatomic molecule when rotational energy is taken 
into account. (Rotation about the axis that joins the two atoms is possible but 
represents a negligible amount of energy.) For molecules containing more than two 
atoms, the number of degrees of freedom of motion is still greater. Kinetic energy of 
vibration is the result of a motion of the atoms of a molecule relative to each other, 
thus changing the pattern of the molecule in rapid periodic fashion. This type of 
motion does not make its appearance in important degree except at the higher tem¬ 
peratures but may account for an additional two degrees of freedom of motion of the 
diatomic molecule. 


170 


BASIC ENGINEERING THERMODYNAMICS 


Let us consider a unit volume of a perfect gas confined within the cubical 
container represented by the dashed lines of the figure; the edges of the 

It will be assumed, to conform to the 
kinetic theory of matter, that the 
number of individual molecules in 
the container is very large, and we 
shall designate that number as N. 
We shall further assume, for the 
present, that only a single perfect gas 
is contained and that the mass m is 
therefore the same for each mole¬ 
cule. Further, it will be assumed 
that the gas system is in tempera¬ 
ture and pressure equilibrium so that 
the absolute velocity of every mole¬ 
cule is the same (although they 
move in different directions at a 
given instant). The immense num¬ 
ber of molecules and the helter- 
skelter nature of their movement 
make it possible to represent their 
average direction as along the diagonal of the cube as indicated by the 
vector V of Fig. 9:1. The components of this average velocity in the X , 
Y, and Z directions will be designated as V x , V y , and V z , respectively. 
These components are equal and, since V 2 = V x 2 + V y 2 + V z 2 , may be 
expressed as V/\/S. 

The pressure exerted against the walls of the container is the effect 
produced by the bombardment of those walls by the perfectly elastic 
molecules. We may calculate the intensity of this pressure if we study 
the impulse exerted by a single molecule, moving in the average direction 
of all of the molecules of the system, against the walls of the container, 
multiplying this single effect by the total of the number of molecules 
in the system. When this average molecule strikes the right vertical 
wall of the cubical container and rebounds, the component of velocity 
V x is reversed and the change in velocity that occurs is 2V X ; neither of the 
components V y or V z is affected since the reaction force applied by the 
wall on the molecule, which causes the molecule to rebound, can be only 
in a direction normal to the wall. The impulse applied on the wall by 
this collision is equal to the change of momentum of the molecule, or 
2 mV x . This molecule will return to the left vertical wall and, rebound¬ 
ing from that wall, will again strike the right wall in the time 2/V x , since 
the perpendicular distance between the walls is unity and the (unreduced) 
velocity in the direction of that perpendicular is V x . The number of 


cube are, of course, of unit length. 



Fig. 9:1. Translational molecular mo¬ 
tion. 










THE PERFECT GAS 


171 


collisions per second is therefore V x /2, and the force exerted against the 
right wall is, accordingly, 

U = 2 mV, ^ = mV, 2 


or, for N molecules and since the side of the cube is of unit area, 


F x = NmVP = P x 


NmV 2 

IT 


To justify our selection of the diagonal of the cube as representing the 
average direction of molecular movement, note that P y and P z are equal 
to P x when this selection is made and that this is a condition necessary 
to the stable equilibrium of the system. P x may therefore be replaced by 
P, the pressure of the gas system. 

The total mass of the molecules of the system is Nm, and this is the 
mass of a system of unit volume so that Nm = p, the mass density of the 
perfect-gas system. Thus the pressure exerted by the system is 

P = ipV 2 


We may substitute l/v for p for this simple system, and thus 

Pv = if 2 (9:4) 

This relation will later prove to be important in the calculation of the 
specific heats of the perfect gas (Art 9:4). 

When two or more gases comprise a mixture, the effects of the bombard¬ 
ments of the walls by each type of molecule may be separated and the 
result called the partial pressures of the constituents. The gases will 
differ in the masses and the velocities of their molecules. Then 

P a = ^ Nam a Va 2 Pb = iN b m h V b 2 etc. (9:5) 

The total pressure of the mixture is the sum of these partial pressures, or 

Pm = Pa J rPb J r'''~\~Pn (9:6) 

Dalton’s law states this relation as follows: The total pressure of a mixture 
of perfect gases is the sum of the partial pressures that would he exerted by the 
constituent gases at mixture temperature and mixture volume. 

When Eqs. (9:1) to (9:3) are combined to allow a variation of all three 
of the primary properties, the primary equation of state of the perfect 
gas results as 

Pv = RT (9:7) 

in which R is a constant, called the gas constant. The gas constant is not 
dimensionless; analysis will show that it has the dimensions of length/ 




172 


BASIC ENGINEERING THERMODYNAMICS 


temperature. If Eq. (9:7) is differentiated with P constant, 

Pdv = RdT (9:8) 

P dv is the work that accompanies a reversible process (in this case, a 
reversible constant-pressure process). Analyzing Eq. (9:8), we see that 
R is the work that is performed during the reversible constant-pressure 
expansion of unit weight of the perfect gas while the temperature increases 
by 1 deg. On this basis, the dimensions of R are work per pound per 
degree Fahrenheit. 

Equation (9:7) has been written for unit weight of the gas. For a 
perfect-gas system of weight M, both sides of that equation may be multi¬ 
plied by the weight, or 

MPv = PV = MRT (9:9) 

The molecular weight m of a substance gives a value to the relative 
weight of its molecule. For instance, m for oxygen is 32 and, for carbon 
dioxide, 44. The mole is defined as an amount of a substance that has a 
weight equal to its molecular weight; the mole of oxygen, as we shall use 
it, will have a weight of 32 pounds (called the pound-mole) although 
other units of weight, such as the gram (the gram-mole, abbreviated 
g-mole), are also used. One mole thus contains the same standard num¬ 
ber of molecules for any gas (or, indeed, any liquid or solid). 

If the substance is a perfect gas, Eq. (9:9) may be applied to a weight 
equal, in pounds, to its molecular weight m, giving 

PV m = mRT (9:10) 

in which V m is the molar volume, or volume of 1 mole. Observing that, 
in Eq. (9:7), R is proportional to the specific volume at given pressure 
and temperature, the product mR in Eq. (9:10) is seen to have the same 
value for any perfect gas. This product, called the universal gas con¬ 
stant, has a value of, approximately, 1545 ft-lb/(°R)(mole). For the 
gram-mole, the corresponding value of the universal gas constant may be 
shown to be 0.848 kg-m/(°K)(g-mole). Note that we shall use no prefix 
to designate the pound-mole. 

When the universal gas constant is substituted for mR in Eq. (9:10), 
the molar volume V m is observed to have the same value for any perfect 
gas when the pressure and temperature are the same. This leads 
directly to Avogadro’s law: At identical temperature and pressure, equal 
volumes of all gases contain the same number of molecules. This statement, 
originally framed to apply to all gases, is exact for perfect gases. Since 
mR = 1545 ft-lb/(°R)(mole), we may calculate the gas constant R for 
any gas as the quotient obtained by dividing the universal gas constant 


THE PERFECT GAS 


173 


by the molecular weight of that gas, or 


R = 


1545 

m 


(9:11) 


Applied to oxygen, this gives a value of R for that gas of = 48.3. 
Equation (9:11) is often particularly useful when the gas system consists 
of a mixture of individual gases. In this case, if the proportions of the 
mixture are known, the weighted average molecular weight (called the 
apparent molecular weight) may be calculated and a value of R obtained 
from Eq. (9:11) to apply to the mixture. Thereafter Eqs. (9:7) and 
(9:9) may be applied to the mixture exactly as if the system had been a 
single gas. Air is, of course, such a mixture; the value of R for air is 
53.3. 

Table 9:1. Approximate Data for Gases 


Name 

Formula 

Mol. wt 

m 

Gas 

constant 

R 

c p * 

Btu/(lb)(°F) 

C v * 

Btu/(lb)(°F) 

k* 

Air. 


28.97 

53.3 

0.240 

0.171 

1.40 

Carbon dioxide. 

C0 2 

44 

35.1 

0.201 

0.156 

1.29 

Carbon monoxide. . . 

CO 

28 

55.2 

0.248 

0.177 

1.40 

Helium. 

He 

4.002 

386.0 

1.25 

0.755 

1.66 

Hydrogen. 

h 2 

2.016 

766.6 

3.42 

2.43 

1.41 

Nitrogen. 

n 2 

28.016 

55.2 

0.248 

0.177 

1.40 

Oxygen. 

o 2 

32 

48.3 

0.217 

0.155 

1.40 

Water. 

h 2 o 

18.016 

85.8 

0.45 

0.34 

1.32 


* Approximate values at atmospheric temperature. 


Table 9:1 tabulates approximate data for certain gases of general 
interest to the engineer. The value of the gas constant has been obtained 
by the use of Eq. (9:11). The specific heats are approximate and are 
those which apply at atmospheric levels of temperature. 


Example 9:2A. Derive Eq. (9:7) from the laws of Boyle and Charles. 

Solution. Let the initial pressure, specific volume, and temperature be represented 
by Pi, Vi, and Ti and the final values of these properties at the end of a certain change 
of state by P 2 , v 2 , and T 2 . The change will be accomplished in two steps, during the 
first of which the pressure is constant at Pi as the temperature changes to its final 
value P 2 ; during the second step, the temperature will be constant at T 2 as the pres¬ 
sure attains its final value P 2 . The specific volume at the end of the first step will be 
denoted as v'. For the first step, applying Eq. (9:2), 


Vl 

v' 


Ti 

= Y °rv 


Vi T 2 
T i 


and, for the second step, making use of Eq. (9:1), 


PiU r = P 2 v 2 or v' 


PzV 2 

Pi 




























174 


BASIC ENGINEERING THERMODYNAMICS 


Equating the two values of v', 

V1T2 P2V2 P1V1 _ P2V2 

~~f\ 0r T\ ~ Ti 

from which Pv/T = a constant = R, or Pv = RT. 

Example 9:2 B. Based on the assumption that it is a perfect gas, calculate (a) the 
density of methane (CH 4 ) at a pressure of 30 psia and temperature of 100°F and (6) the 
molar volume at the same pressure and temperature. 

Solution. The molecular weight of methane is 12 + (4) (1.008) = 16.032. 


(a) 


R = 


1545 

16.032 


96.4 


1 _ P (30) (144) 

w ~ v RT (96.4)(560) 

_ 1545T (1545) (560) 

m “ P (30)(144) 


0.08 lb/ft 3 
200 ft 3 


Example 9:2 C. Show that R will always be the maximum work that can accompany 
the constant-pressure expansion of a perfect-gas system of unit weight as the tempera¬ 
ture increases by 1 deg, without regard for the initial pressure or temperature that is 
assumed. 

Solution. The maximum work is P(v 2 — Vi). 


Vi = 


RTi 

Pi 


•; V2 = 


RT 2 R(T x 4- 1) 
2 Pi 


(Ti + 1 ,\ 1 

; V2 - VI = VI ( -yT-1 ) = Vi Y 1 


P(.V, - »,) = Pm ± ^ - R 


9:3. Joule’s law states: The internal energy of a gas is independent of the 
volume and depends only on the temperature of the gas. Joule arrived at 
this conclusion as the result of a famous series of experiments. Two 
vessels were connected by a pipe and valve and immersed in a bath of 
water. One of the vessels contained air at high pressure; the other was 
evacuated. When the valve was opened and the pressure was equalized 
between the vessels, Joule observed no change in air or water tempera¬ 
ture. No external work had been performed, and the constant tempera¬ 
ture of the water bath indicated that there had been no heat flow; he 
therefore concluded that there had been no change of internal energy 
and stated his law accordingly. Later investigation, based on more 
critical measurements and more refined apparatus, showed that all real 
gases deviate at least slightly from the behavior which Joule observed; 
the deviation for gases which, like air and hydrogen, are not easily 
liquefied, is very small. This deviation is expressed in terms of the 


Joule coefficient which is defined as g 



; its similarity to the 


Joule-Thomson coefficient, g J} will be noted. The first applies to a non¬ 
flow process, the second to a process involving continuous flow. 

It may be shown that Joule’s law will hold exactly for the perfect gas. 















THE PERFECT GAS 


175 


The perfect gas is defined by the primary equation of state, Pv = RT. 
This equation conforms to Eq. (1:6) when R has the value 144C, and 
Eq. (1:6) is therefore one form of the primary equation of state of the 
perfect gas. It was shown in Chap. 8 (Example 8:2 B) that the internal 
energy of this gas (now recognized as a perfect gas) is a function of the 
temperature alone and is independent of the volume. The general 


definition of the specific heat at constant volume as c v = 



may 


therefore be changed, as it applies to a perfect gas, to the form 


Cy 


du 

df 


(perfect gas) 


( 9 : 12 ) 


Moreover, it was additionally proved in Example 8:2 B that the 
enthalpy of the perfect gas is also, like the internal energy, a function 


of the temperature alone. The general definition of c p as 


(A) 

Wi 


there¬ 


fore becomes, for the perfect gas, 

dh 


Cp ciT 


(9:13) 


Reference to Eq. (9:7) indicates 
that a line of constant temperature 
as drawn on a PV chart for any per¬ 
fect-gas system would correspond to 
a locus for which the product of pres¬ 
sure and volume is constant and 
would therefore appear as a hyper¬ 
bola. But it has been shown that 
all points along this line of constant 
temperature also represent states of 
equal internal energy and equal en¬ 
thalpy. This is illustrated in Fig. 



Fig. 9:2. Internal energy and enthalpy 
changes for a perfect gas. 


9:2; the lines of respectively constant temperature T x and T 2 are also 
labeled Ui and U 2 and Hi and H 2 , to indicate that these properties are 
also the same for all points along them. The line ab represents a con¬ 
stant volume process connecting points on the two lines; the change of 
internal energy that accompanies this process is 


U t - U a = U t - I/, = M ff c. 


dT 


(9:14) 


Turning now to the process represented by cd, a constant-pressure process, 
we observe that the change of internal energy (U d — U c = U 2 — U i) is 
the same as for constant-volume process ab and therefore is measured by 









176 


BASIC ENGINEERING THERMODYNAMICS 


applying Eq. (9:14). This could be the case only if c v is either constant 
or a function of the temperature alone. This has been shown to be the 
case in Example 8:2 B. Continuing, we may show by a similar reasoning 
procedure that, for any process that begins at temperature T\ and ends 
at temperature I\ (such as process ef of the figure), Eq. (9:14) will 
furnish a means of evaluating the change of internal energy. 

Similarly, it may be shown that the change of enthalpy for constant- 
pressure process cd is 

H d - H c = H 2 - H, = M C c p dT (9:15) 

J T i 


and that this expression will also measure the change of enthalpy of the 
system over any process (such as ab or ef) that begins at temperature 
T i and ends at temperature T 2 . 

It has been shown above [see Eq. (9:8) et seg.] that R measures the work 
performed as a closed system of unit weight expands reversibly at con- 

(dll )p, rev 


stant pressure owing to a unit increase of temperature, or R = 


dT 


The accompanying change of internal energy, du, is, from Eq. (9:14), 
equal to c v dT for a perfect-gas system, and the heat flow, ( dQ) Ptiev , is, 
differentiating Eq. (2:9), equal to dh and therefore to c p dT. Substitut¬ 
ing these values in Eq. (2:5), we obtain 


c p dT — ^ dT = c v dT 


or 


R 



(9:16) 


From this relation and the ratio of the specific heats ( k = c p /c v ), other 
useful relations may be obtained, as follows: 


kR 

p ~ J{k - 1) 
R 

J(k - 1) 


(9:17) 

(9:18) 


Example 9:34. During a certain adiabatic process, an air system weighing 3 lb 
changes from p i = 100 psia, t x = 200°F, to p 2 = 20 psia, t 2 = —43°F. Assuming 
air to be a perfect gas, calculate (a) the change of internal energy of the system, ( b) the 
change of enthalpy, and (c) the external work that accompanied the process. 

Solution: 


(a) U 2 - Ux = Mc v (T 2 - T\) = (3)(0.171)(417 - 660) = -124.7 Btu 

(b) H 2 — Hi = Mc p (T 2 - Ti) = k(U 2 - Ui) = (1.4)(-124.7) = -174.6 Btu 

(c) iQ 2 = 0 = C/ 2 - Ux + ^y- 2 or AV 2 = -J(U 2 - Ui) = (-778) (-124.7) 

= 97,000 ft-lb 

Example 9:3 B. Calculate c p and c v for methane if k = 1.31 and it is assumed to 
be a perfect gas. 






THE PREFECT GAS 


177 


Solution. In Example 9:2R, R was calculated for methane as 96.4. 


Cp 


Cy 


kR 

J (k — 1) 
c p = 0.525 
k ~ 1.31 


(778)(0.31) 0.525 Btu/(lb)(°F) 

= 0.401 Btu/(lb)(°F) 


9:4. The Ratio of the Specific Heats. Equation (9-16) indicates that 
the difference between the specific heats of a perfect gas is constant. The 
ratio of the specific heats, k , can therefore be constant only if c v and c v 
are themselves constant. That this is the case for the monatomic perfect 
gas, where molecular motion is translational only, is proved below. 

From Eq. (9:4), 



and, from Eq. (9:7), 
Therefore 


The kinetic molecular energy is, for unit mass, V 2 /2. This is the total 
internal energy of the perfect-gas system of unit mass, or 



V 2 3 RT 

U 2J 2 J 

(9:19) 

Differentiating, 

du 3 R 

Cv ~ dT ~ 2 J 

(9:20) 

From Eq. (9:16), 

R 5 R 
c„- c* + J - 2 J 

(9:21) 

and 

7 Cp O 

k = Z = 3, 

(9:22) 


The value of c v stated in Eq. (9:20) and the value of k given in Eq. 
(9:22) are approached by all gases, whether monatomic or polyatomic, as 
their temperature approaches absolute zero, since, at these levels of tem¬ 
peratures, rotational and vibrational kinetic energy vanish. At slightly 
higher temperatures, rotational energy begins to appear and, at tempera¬ 
tures still well below atmospheric levels, reaches its maximum amount. 
When this condition has been attained, the rate of storage of internal 
energy with respect to temperature is increased above that due to changes 


Pv = RT 
V 2 


3 


= RT 









178 


BASIC ENGINEERING THERMODYNAMICS 


in translational kinetic energy alone and stated in Eq. (9:20). It is a 
principle of quantum mechanics that the total kinetic energy will be in 
direct proportion to the number of degrees of freedom of motion of the 
system (in this case, the molecule). When the rotational kinetic energy 
of a two-atom (diatomic) molecule has been fully excited, the molecule 
adds two additional degrees of freedom of motion to its three degrees of 
freedom of translational motion, since it may store energy in significant 
amount due to rotation about either of two axes perpendicular to each 
other and to the line joining the two atoms. This makes a total of five 
degrees of freedom of motion, and the rate of change of internal energy 
of the perfect gas with respect to temperature in this range of temperature 
(before vibrational kinetic energy makes its appearance in appreciable 
degree) is five-thirds as large as when translational kinetic energy was 
alone involved. Thus, for the diatomic perfect gas in this range, 


n 



5 R 
2 J 


Cp 


R = 7R 
J 2 J 



\ (9:23) 


At temperatures approximating those of the atmosphere, vibrational 
kinetic energy begins to make its appearance in appreciable amount. 
This motion accounts for still two more degrees of freedom, and, for the 
diatomic molecule in which vibrational movement of the atoms has been 


fully developed, 

7 R 

C ° = 2 J 


_ 9 R 
Cp ~ 2 J 



(9:24) 


Equations (9:24) state the highest values of c p and c v (and the lowest 
value of k) theoretically possible for the diatomic perfect-gas molecule. 
The change from the values shown in Eqs. (9:23) to those of Eqs. (9:24) 
is gradual and begins to make its appearance at about atmospheric 
temperature. Calculations based on spectroscopic data for real dia¬ 
tomic gases, 1 such as 0 2 and N 2 , seems to show that c p may, at tempera¬ 
tures around 9000°R, somewhat exceed the maximum as stated in Eq. 
(9:24) if R is calculated as for a perfect gas. 

If it is assumed that vibrational kinetic energy has not yet made its 
appearance, a triatomic molecule may have three degrees of rotational 
in addition to three degrees of translational freedom. Thus, at about 
atmospheric levels of temperature, we should expect to find 

6 R SR ,8 

c * = 2 J Cp = 2 J ^ = - = L33 (9:25) 

Observed values of k at these temperature levels are often somewhat less 

1 R. L. Sweigert and M. W. Beardsley, “ Empirical Specific Heat Equations Based on 
Spectroscopic Data,” Georgia School Technol, State Eng. Expt. Sta., Bull. 1 (No. 3) 
(1938). 



THE PERFECT GAS 


179 


than this theoretical maximum for triatomic gases such as C0 2 and H 2 0, 
and, ignoring the difference between these gases and their perfect-gas 
equivalents, we may account for this variation as due to the accumulation 
of vibrational energy in some degree, even at atmospheric temperature. 

It will be remembered (see Art. 9:1) that the concept of the perfect 
gas was introduced to simplify the calculation of the effects produced 
during the processes which make up the gas cycle. These calculations 
then would be, in effect, approximate calculations to apply to the real 
gas. They may be still further simplified if it is assumed that the 
specific heats of the perfect gas, which is substituted for the real gas, are 
constant. This will be very nearly the case over the rather wide range 
of temperature between the temperature at which rotational energy 
is fully developed and that at which vibrational energy begins to be 
important; this second level starts at about atmospheric temperature. 
Accordingly, we shall use, as constants, the values of the specific heats 
at atmospheric levels of temperature in the calculations to follow. 

Example 9:4. (a) Helium is a monatomic gas, having a molecular weight of 4.002. 

Assuming it to be a perfect gas, calculate c v and c v . ( b ) Check the values of c p and c v 
shown in Table 9:1 for carbon monoxide, assuming it to be a perfect gas. 

Solution: 


(a) As a monatomic gas, the value of k is § for helium. 


kR _ (5) (1545) 

Cp “ J(k - 1) (778) (2) (4.002) 

c„ = ^ = 0.745 Btu/(lb)(°F) 


1.24 Btu/(lb)(°F) 


(6) The tabulated values are at atmospheric temperature. This is a diatomic gas, 
and, assuming that rotational energy is fully developed and vibrational energy is 
negligible, k = \ [see Eq. (9:23)]. Then 


Cn — 


Cv 7 


17 - SS) " °' 248 Btu/(lb)( ° F) 

7 = = 0.177 Btu/(lb)(°F) 

k 1.4 


9:5. Reversible Nonflow Processes. It is customary to classify 
reversible processes for the perfect gas according to the value of the 
constant exponent n in the pressure-volume relation 

PV n = C (9:26) 


This is done since, as will develop later, if c v (and, consequently, c p ) for 
the perfect gas is constant, the amount of heat flow that accompanies a 
unit change of temperature is the same over every stage of the process. 
This also accounts for the name, polytropic process , which is applied to 
any reversible process of the perfect gas that can be expressed in the 
form of Eq. (9:26); the adjective polytropic, freely translated, means “at 






180 


BASIC ENGINEERING THERMODYNAMICS 


constant specific heat.” To symbolize this constant specific heat we 


shall use the character c„, representing the heat flow in Btu that accom¬ 
panies an increase in temperature of 1°F per pound of system weight. 

From Eq. (2:3), i Q 2 — 1 W 2 /J = U 2 — U\. Assuming, for the 
moment, that c n is constant during the polytropic perfect-gas process 
to which we expect to apply this equation, 


(9:27) 


1 Q 2 — Mc n {T 2 — T 1 ) 

Based on Eq. (2:6), and since P = C/V 11 in Eq. (9:26), 




—n _ 


or since C = PV n = PiVp = P 2 V 2 n , 


P 2 F 2 n F 2 1 - n - PiV\ n Vi l ~ n 

1 — n 


P 1 V 1 - P 2 V 2 

n — 1 


(9:28) 


Alternately, since P\V\ — MRT 1 and P 2 F 2 = MRT 2 , if the system is a 
perfect gas, 



(9:29) 


From Eq. (9:14), if c v is constant, 


lh - lh = Mc v (T 2 ~ T 1 ) 


(9:30) 


Substituting the expressions obtained from Eqs. (9:27), (9:29), and (9:30) 
in Eq. (2:3) and rearranging, 



or, dividing by M(T 2 — T 1 ) and applying Eq. (9:18), 




Since n is constant by definition and k is constant if, as was assumed, c v 
and c p are constants, c n is a constant and the term polytropic, as applied 
to the process, is justified. 

The relations between the heat flow, the change of internal energy, and 


the external work of a perfect-gas system during a reversible polytropic 


process may be obtained by comparing their values as expressed in 
Eqs. (9:27) to (9:30). For example, 


1 Q 2 ??/ k 


C 2 U 1 Cy 


n — 1 


(9:33) 















THE PERFECT GAS 


181 


Similarly, 


and 


J 1 Q 2 Jc n (n — 1) 


(n - k)(k - 1) 


lW 2 —R 71—1 

J (U 2 — U i) Jc v (n — 1) 7i — l 


(9:34) 


(9:35) 


1 W 2 ~R k — 1 

For the polytropic, dQ is Mc n dT [the differential form of Eq. (9:27)], and 

,' dQ __ [T*dT ,, . P 2 n — k. T 2 

S 2 — Si — / p Mc n I rp Mc n loge ^ J\IC v ^ ^ log e p 


Ti 


(9:36) 


The reversible polytropic is the general form which fits all perfect gases 
with constant c v to give a process for which the ratio of heat flow to 
temperature change is constant. The reversible constant-pressure, 
constant-volume, isothermal, constant-internal-energy, and adiabatic 
processes discussed in Chap. 2 are special cases of the polytropic, each 
distinguished by a special value of the constant exponent n and a cor¬ 
responding constant value of c n . For the constant-pressure process, 
P = C, which may be written as PV° = C, and the value of n is therefore 
zero. The constant-volume process, V = C, becomes P°V — C, or 
pi /oo F = C, and, raising this to the infinite power, we have PF 00 = C°°, 
where C 00 is still a constant; n for the constant-volume process is 00. 
The isothermal, or constant-temperature, process follows Boyle’s law 
and is stated as PV = C, so that n is 1; from Joule’s law, the same value 
of n will also apply to the constant-internal-energy process. The 
reversible adiabatic involves no heat flow, and c n is therefore zero. But 
c n = c v (n — k)/(n — 1) and since c v cannot be zero (except at the 
absolute zero of temperature), (n — k)/(n — 1) =0, or n = k = c p /c v . 
The equation of the reversible adiabatic for the perfect-gas system is 
therefore 

PV k = C (9:37) 


Example 9:5^4. Show, by applying the nonflow energy equation to a perfect gas, 
that n = k for a reversible adiabatic (isentropic) process. 

Solution: 


dQ 


dW 

J 


= dU 


But dQ = 0 if the process is adiabatic, and, if reversible, dW = PdV. 


[Eq. (2:5)] 
Thus 


0 = dU + 


PdV 

J 


Mc v dT + 


PdV 

J 


For a perfect gas, PV = MRT, and, differentiating, P dV + V dP = MR dT. Solv¬ 
ing for dT = + -ryw and substituting in the preceding equation, 

IVl it iVJ. it 

















182 


BASIC ENGINEERING THERMODYNAMICS 


c v P dV . c v VdP . PdV _ 

~r ~ + ~ + ~r _ 0 

or, assembling coefficients of dF, 

(1+7)^^ + ® VdP = 0 

Multiplying by R, 

(c v + P dV + c„F dP = c p P dF + c v V dP = 0 

Separating the variables by dividing by PF, 

dF , dP _ kdV x dP _ 
y' ^ V ' P ^ 

Integrating, 

k loge F + log c P = C (constant of integration) 

This relation may be expressed as PV k = C, where C is a new constant equal to PiFi*. 

The family of polytropics may be traced on a PV chart as in Fig. 9:3. 

For positive values of n, the curves 
lie in the second and fourth quad¬ 
rants, if the common initial state 1 
for all processes is used as the 
origin, and these quadrants include 
all the more usual processes. 
When the rate of heat and/or work 
flow is large enough so that the 
pressure and volume may increase 
or decrease together (a condition 
seldom encountered in practice), 
the value of n will be negative and 
the curves will lie in the first and 
third quadrants. 

The equation of state Pv = RT 



Fig. 9:3. Polytropic processes on the PF 
chart. 


and the poly tropic relation Pv n = C may be used to determine the PvT 
relationships between two states that lie on the same polytropic. From 
the first, 


P1V1 _ P2V2 
T x T 2 


or 


T2 

T x 


P2V2 
PlV 1 


(9:38) 


and, from the second, 















THE PERFECT GAS 


183 


Combining Eqs. (9:38) and (9:39), 


Ti 
T ! 



n —1 


or 


Tj 

T 


2 = ^y n - i)/n 


(9:40) 


The appearance of the perfect-gas polytropics on the temperature- 
entropy diagram is shown in Fig. 9:4. A discussion of the method of 
location of the constant-volume, constant-pressure, isothermal, and 
reversible adiabatic paths on this 
diagram will be found in Chap. 6. 

The direction 1-2 on these curves 
corresponds to increasing volume or, 
in the case of the constant-volume 
path, to decreasing pressure. 

The substitution of the proper 
value of n in Eqs. (9:26) to (9:40) 
leads directly to the information 
tabulated in Table 9:2 except in the 
case of the isothermal process; for 
this process, the heat flow, the work, 
and the change of entropy are inde¬ 
terminate from these equations. 

But the change of internal energy 
accompanying this process is zero for 
a perfect gas according to Joule’s law, and, by Eq. (2:3), the heat flow 
must equal the work of the reversible process. The equation of the path 
is py = C = PiVi = MRT, or P = C/V = P{Vi/V = MRT/V. Thus 
the work may be calculated as 

f 2 f V2 dV Vo Vo 

iTF 2 = / P dV = P 1 V 1 / = P 1 V 1 log* ^ = MRT log* ^ (9:41) 

and the heat flow is 



Fig. 9:4. Polytropic processes on the 
TS chart. 


1Q2 — 


iTF 2 P.Vi 


J 


J 


, V 2 MRT. V 2 
l0g 'V = 


(9:42) 


This heat flow is represented by the rectangular area under the n = 1 
curve of Fig. 9:4, and the change of entropy is therefore 



1Q2 

~Y 


PiV 1 , f 2 
JT oge F 1 


MR, V 2 

— l0g ‘V 


(9:43) 


Table 9:2 will be found convenient for reference in the solution of prob¬ 
lems concerned with nonflow reversible processes of perfect-gas systems 
for which c v has a constant value. 


















Table 9:2. Summary of Nonflow Reversible Processes for Perfect-gas Systems 

c v constant 


184 


BASIC ENGINEERING THERMODYNAMICS 


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* MRT may be substituted for PiFi or P1/P2 for F2/F1 if more convenient, 
t (Fi/F2 ) n_1 or (Pi/P2) (n_1) / n may be substituted for T2/T1 if more convenient. 








































































THE PERFECT GAS 


185 


Example 9:5 B. As the result of a polytropic (reversible) process, a 3-lb air system 
changes from 100 psia, 200°b, to 20 psia, 0°F. Calculate (a) the change of internal 
energy, ( b ) the heat flow, (c) the external work, ( d ) the change of enthalpy, and (e) the 
change of entropy that accompany the process. Consider the system to be a perfect 
gas with constant specific heats, c v and c v . 

Solution: 


(a) U 2 - Ui = Mc v (T 2 - TJ = (3) (0.171) (460 - 660) = -102.6 Btu 
0 b ) From Eq. (9:40), 


r 2 

(P 2 V* 

■—1 )/n 

460 / 

Ti " 

' \Pi) 


U1 660 ~ ' 

Thus 




n — 

1 _ log 

4 6 0 
6 6 0 

-0.362 

n 

log 

2 0 
ToTT 

-1.61 


/ 20 \(»—D/n 

= vW 


= 0.225 and n = 1.29 


- *;hri - 0-171 ( ‘g 9 _\ 4 ) - -0 065 

,Q> = Mc„{T 2 - Ti) = (3) (— 0.065) (— 200) = 39.0 Btu 


_ MR(Ti - T t ) (3)(53.3)(200) 

<C> ' Wl -- = - 5^9 - 


110,000 ft-lb 


Checking, 

iW 2 = J(iQ 2 ~ A U) = 778(39.0 + 102.6) = 110,000 ft-lb 

(d) H 2 — Hi = Mc p (T 2 ~ Ti) = (3)(0.240) (-200) = -144 Btu 

(e) S 2 - Si = Mc n log,?? = (3)(-0.065) log e ^ = (3)(-0.065)(-0.362) =0.071 

9:6. The Irreversible Nonflow Process. Even if n has a constant 
value in the equation PV n = C, that equation does not necessarily repre¬ 
sent a polytropic unless the process is a limiting (reversible) process, and 
the information contained in Table 9:2 cannot be applied to the calcula¬ 
tion of either the heat flow or the work that accompanies the process. 
Heat and work have been shown to have maximum values (in the alge¬ 
braic sense) for the reversible process. As applied to the adiabatic 
process, for example, the reversible process has been shown to have a 
value of n equal to the ratio of the specific heats, or k. If, owing to the 
effects of factors such as friction or unrestrained expansion, the work 
accompanying an adiabatic expansion were reduced to zero (see Chap. 2), 
the process would take place at constant internal energy and n would be 
1 for the perfect-gas system. Even this is not a limiting case, for it is 
conceivable that the work might even be negative and n thus less than 1. 
The value of n during an irreversible adiabatic for a perfect gas must be 
less than k, the deviation from that value depending on the amount of 
irreversibility. 

Although a constant value of n during an irreversible process for a per¬ 
fect gas does not stamp the process as a polytropic or fix the values of work 











186 


BASIC ENGINEERING THERMODYNAMICS 


or heat flow, it does establish the series of states through which the system 
has passed and the corresponding change in properties such as the internal 
energy, enthalpy, and entropy. Also, although their individual values 
are not calculable, the difference between the heat flow and the work may 
be determined. This is true, of course, even if n is not constant over the 
path connecting the initial and final states of the system. In fact, the 
path need not be known, and the calculation may be based on the end 
states alone. If the path is known, limiting values of the work and the 
heat flow are established and these values cannot be exceeded for any 
process that follows the same path. 


Example 9:6. Assume, in Example 9:5 B, that the same system follows the same 
path but as the result of an irreversible adiabatic process. Find the same quantities 
as in Example 9:56. 

Solution. The internal energy, the enthalpy, and the entropy are properties, and 
their changes depend only on the end states, which are the same for the irreversible 
as for the reversible process of Example 9:5.6. The answers to parts a , d, and e are 
therefore the same as in that example. The heat flow is, according to the description 
of the irreversible process as an adiabatic, zero. The work may be calculated from 
Eq. (2:3) as 

iW 2 = JGQt - AU) = 778[0 - (-102.6)] = 79,800 ft-lb 

It will be noted that this amount of work is less than the maximum work (110,000 
ft-lb) that is associated with the reversible process. Also observe that the value of 
n (1.29) is less than k for this gas. 


9:7. Changes of Entropy. In Table 9:2 are given the changes of 
entropy over reversible polytropic processes connecting an initial state 1 
with a final state 2. It was shown in Chap. 6 that this change of entropy 
was the same for the same end states regardless of the process, reversible 


over 


or irreversible, that connected them and could be calculated as f — 

J i ^ 

any reversible process or processes that joined the two states; this was the 
basis of Eq. (6:8). Starting with that equation, we may write, for the 
perfect-gas system, 


dS 


dU PdV 
T + JT 


Mc v dT 

+ 

MR dV 

(9:44) 

T 

JV 

Mc v d(PV) 
PV 

MR dV 
+ JV 

(9:45) 

Mc p dV 
V 

+ 

Mc v dP 

P 

(9:46) 

Mc p dT 


MR dP 

(9:47) 

T 


JP 


Integration between the limits determined by the properties at states 1 
and 2 gives, respectively, 













THE PERFECT GAS 

187 

m i T 2 . MR . V 2 

- MC V l0g e + j loge y 

(9:44a) 

^ , PM 2 MR , F 2 

- Mc v log e p iVi + J °ge y 

(9:45a) 

y p ^ 

= Mc p log c j7 - 2 + Mc v loge M 

V i r i 

(9:46a) 

hj i T 2 M R . P 2 

= Mc p loge rji - J loge p 

(9:47a) 


Equations (9:44a) to (9:47a) will be found convenient when the differ¬ 
ence of entropy between two known states is to be calculated without any 
knowledge of what single reversible polytropic would connect these states. 
They assume that the change of state is accomplished in two reversible 
steps and add the individual changes of entropy of these two segments to 
obtain the total change. For example, in Eq. (9:44a), the two segments 
are, respectively, at constant volume and at constant temperature. An 
alternate method would consist in solving for n in Eq. (9:39) or (9:40) and 
then applying the appropriate equation from Table 9:2 as in Example 
9 :5B. 

Example 9:7. Calculate the change of entropy in Example 9 :5B by selecting and 
applying a suitable equation from Eqs. (9:44a) to (9:47a). Compare with the value 
previously calculated. 

Solution. The initial and final pressures and temperatures being given, it will be 
convenient to make use of Eq. (9:47a). Then 

(3) (53.3) (-1.61) 
778 


S 2 - S, = Mc„ log. ~ - 

i 1 


MR 


J log.y = (3) (0.240) (-0.362) - 


= -0.261 + 0.332 = 0.071 

This agrees with the value as calculated in Example 9:5 B. 

9:8. The Steady-flow Process. Equation (3:5) may be written 

i W s 


\Qz 


J 


2 , . , A(F 2 ) A* 

= hi - hl + ~2jf + T 


and, for the change of enthalpy, h 2 — h h as it applies to a perfect gas, we 
may substitute 

h i -h 1 = c p (T 2 - 7\) = k{u 2 - mi) = h Vt ~ PlVl 


J 


kP iV] 


~(p 2 y- i)/n 


- 1 


(i - dj ■ - ■ - (9:48 » 

If the steady-flow process is reversible and adiabatic, with negligible 
changes in kinetic energy and elevation, n — k and 

(fc—1)/A;' 


iTFa = (hi - h 2 )J = 


k 


k - 1 


PlVl 


- 


(9:49) 



















188 


BASIC ENGINEERING THERMODYNAMICS 


To deal with the more general case where the steady-flow process, 
though poly tropic (reversible with n constant), is not adiabatic, we may 
write the energy equation for steady flow, again assuming negligible 
change in kinetic energy and elevation, in differential form as 



and note that dW = — v dP. For the poly tropic, Pv n = C = P ivY 
= P 2 v 2 n , and v = PA /n Vi/P l/n = P 2 l/n v 2 /P l/n . Therefore, for the reversi¬ 
ble steady-flow polytropic process, 

[2 fP 2 

( 1 TF 2 ) 


rev. steady flow 


VdP- -PiV'Vi J ^ 


n 


n — 1 


(P1V1 - P2V2) = 


= P T /n v ] 

n 


n 


n — 1 


P1V1 


n — 1 
1 - 


-ffi) 


(Pl(»-l)/n p^(n— 1)/«) 

(»—!)/»" 


(9:50) 


It will be noted that Eq. (9:49) is a special case of Eq. (9:50). 

When the steady-flow process is isothermal, n — 1 and Eq. (9:50) yields 
an indeterminate result. Returning to dW = — v dP, however, and sub¬ 
stituting v = P 1 V 1 /P, we obtain, for the reversible isothermal steady-flow 
process, 


1W2 



—P1V1 



= PiVi loge 


Pi 

P <1 


(9:51) 


9:9. The continuous compression of gases is frequently necessary in 
the course of engineering operations. This may be accomplished by 
apparatus such as fans, blowers, or reciprocating compressors, the chief 
thermodynamic distinction between these devices being found in the 
differential of pressure which they create. In general, the reciprocating 
compressor is associated with the larger differentials of pressure and will 
therefore be used as the basis of our discussion. Although the flow is 
quasi-steady (see Art. 3:9) to at least some degree in all these devices, 
Eqs. (9:49) to (9:51) may be used to obtain the work required for the com¬ 
pression of unit weight if an equivalent perfect gas is substituted for the 
real gas and if the processes making up the cycle of operation are idealized 
as reversible processes. The equations will show this work to be nega¬ 
tive, indicating that it must be supplied by some external agency. By 
taking into account the rate at which the gas is compressed, the ideal 
(minimum, or theoretical) power required to drive the compressor may be 
calculated. 

The principle of operation of the reciprocating air compressor is shown 
in Figs. 9:5 and 9:6. The compressor is shown, for simplicity, as single- 













THE PERFECT GAS 


189 


acting although valves at both ends of the cylinder would approximately 
double its capacity at the same speed. Both intake and exhaust valves 
are usually automatic in their action, being opened and closed by differ¬ 
ences in the air pressure acting on their opposite faces; when the pressures 
are balanced, they are kept closed by light springs (not shown in the 
figure). In operation, the pressure of the atmosphere acts on the outside 
of the intake valve; that of the air at receiver (discharge) pressure on the 
outside of the discharge valve. Thus, if the pressure within the cylinder 
is caused to fall below atmospheric pressure by an amount sufficient to 
overcome the small resistance offered by the valve spring plus the resist¬ 
ance to flow offered by the intake passage, the intake valve will open and 




Fig. 9:5. Piston air compressor. Fig. 9:6. Cycle of an air compressor. 

air will be drawn from the atmosphere into the cylinder. Similarly, if, as 
the result of piston action, the pressure of the charge within the cylinder 
reaches a value sufficiently above the receiver pressure to overcome the 
resistance of the spring on the discharge valve plus exhaust-passage 
resistance, this valve will open and discharge will take place from the 
cylinder into the receiver. 

In the position shown, the piston is moving to the right on its suction 
stroke. Referring to the PV diagram (Fig. 9:6), which has been drawn 
for a compressor with no clearance, a charge of air will be drawn into the 
cylinder at constant pressure Pi; this pressure, as explained above, will 
be slightly below that of the atmosphere from which the flow takes place, 
and atmospheric pressure is shown as a dashed line on the diagram. 

At point 1, the piston reaches the end of its suction stroke and reverses 
the direction of its motion. The intake valve closes immediately, since 
the pressure in the cylinder is no longer low enough to keep it open, and 
compression of the charge proceeds along the line 1-2. At point 2, the 
pressure P 2 within the cylinder has become slightly higher than the 
receiver pressure, and this differential of pressure is sufficient to open the 
discharge valve; the remainder of the stroke of the piston as it continues 
to move to the left is concerned with the delivery of the compressed charge 








































190 


BASIC ENGINEERING THERMODYNAMICS 


to the receiver. As the piston reaches the left end of the cylinder and 
again starts on its suction stroke, it will be observed that the discharge 
valve will close, the intake valve will open, in response to the lowered 
pressure within the cylinder, and the cycle of operation of the compressor 
will have been completed. 

Figure 9:6 is a theoretical indicator diagram for the air compressor with 
no clearance. The drop in pressure from P a to P\ across the intake valve 
is the effect of a throttling process, adiabatic (though not frictionless and 
therefore not isentropic) and with no external work. Considering the 
differences of kinetic energy and elevation to be negligible, hi = h a or, 
handling the air as a perfect gas, Ti = T a . The enclosed area of this dia¬ 
gram measures the work required for the cyclic compression of a unit 
weight of the gas if the volume at point 1 is V\, the specific volume at Pi 
and Ti{ = T a ). As indicated on the figure, this area is the summation of 
the slices V dP (for unit weight, vdP), and since Eqs. (9:49) to (9:51) 
represent the result of this summation, these equations may be used to 
measure the work area. The particular equation selected for the purpose 
will depend on the value of n over the compression curve 1-2. 

If the compression occurs rapidly in an insulated cylinder, it may be 
assumed to be adiabatic, with n = k; if carried out very slowly, with 
proper provision, such as a water jacket, for the removal of heat from the 
cylinder, it will approach the isothermal, with n — 1. If it is assumed 
that compression begins from atmospheric temperature and that the tem¬ 
perature of the fluid used to cool the cylinder is not below that of the 
atmosphere, these two values of n may be regarded as the limiting values 
for the reversible compression. The value of n for a water-cooled air 
compressor will be a compromise between these values and may be 
expected to lie in the range 1.25 to 1.35. This would correspond, in the 
compression of a triatomic gas such as carbon dioxide, with k = 1.29, to a 
practical value of n of about 1.2. In the case of an uncooled compressor, 
such as a blower, the value of n will exceed k owing to irreversibilities. 
The exponent n is assumed to be constant along the curve 1-2 because of 
the simplicity of calculation thereby made possible; this does not corre¬ 
spond to the thermodynamic probabilities since, with the difference 
between cooling-fluid temperature and charge temperature a minimum 
at the beginning of compression, the heat flow should be very slow during 
the early stages of process 1-2, with n approaching k. Further along, as 
the temperature differential increased, the rate of heat flow should 
increase and n should correspondingly decrease. 

Figure 9:7 illustrates the saving of work that results from cooling the 
compressor cylinder. Curve A is a PV k = C relation and represents 
adiabatic compression of the charge, curve B represents isothermal com¬ 
pression, and curve C is the usual compromise between these limiting 


THE PERFECT GAS 


191 


values. The respective cyclic areas show the effect of cooling in reducing 
the work of cyclic compression. It will be noted that the proportional 
reduction in this area for isothermal 
as compared with adiabatic compres¬ 
sion increases very rapidly with the 
pressure ratio, P 2 /Pi. When this pres¬ 
sure ratio is low, as for fans and 
blowers creating relatively small pres¬ 
sure differentials, cooling becomes less 
important. 



Fig. 9:7. Comparison of adiabatic 
and isothermal air-compressor dia¬ 
grams. 


Example 9:9. Calculate the power theo¬ 
retically required for the compression of 100 
lb of air per minute from 14 psia and 70°F to 
a final pressure of 350 psia when the compres¬ 
sion is (a) adiabatic, (6) isothermal, and (c) when n = 1.3. 

Solution. For P\V\ in Eqs. (9:49) to (9:51) may be substituted RT\. 

(a) For adiabatic compression, using Eq. (9:49), 

k 


Work per min = 


k - 1 


MRTy 


■ - (£) ] 


= 9,900,000(1 - 2.51) = -14,950,000 ft-lb/min, or 453 hp 

( b ) For isothermal compression [Eq. (9:51)], 

Work per min = MRT\ log e ^ = (100) (53.3) (530) log e 

= (2,830,000)(-3.22) = -8,830,000 ft-lb/min, or 267 hp 

(c) For polytropic compression with n = 1.3 [Eq. (9:50)], 

'Po\ D/» 


Work per min = 


n 

n — 1 


MRTr 


= (g|) (100) (53.3) (530) 


l _ (350y 3/l s 
= 12,250,000(1 - 2.105) = -13,500,000 ft-lb/min, or 410 hp 

For fans and blowers, because of the relatively low pressure differential, 
v 2 is only very slightly smaller than V\. If their cycle of operation is 
placed on a PV diagram, the enclosed area is observed to be nearly 
rectangular and equal, approximately, to (P 2 — Pi)V. Thus it has 
become customary to use the following simplified expression for the theo¬ 
retical power input to such devices: 


hp = 


(P 2 - Pi) 7 _ Az V _ M Ag 


33,000 


33,000^ 33,000 


(9:52) 

















192 


BASIC ENGINEERING THERMODYNAMICS 


in which V = volume of gas handled per minute, ft 3 
P 2 — P 1 = differential of pressure created, psf 

Az = head of gas equivalent to this pressure difference, ft of gas 
v = specific volume of gas, ft 3 /lb 
M = weight of gas handled, lb/min 

If Az is calculated as based on the initial pressure and temperature, the 
result will be slightly too large and, if based on discharge density, slight!) 
too small; if the differential of pressure is less than 3 per cent of Pi, the 
error is usually considered negligible. 

9:10. Multistage Compression. If the ratio P 2 /P 1 exceeds. 5, for 
example, if air is compressed from atmospheric pressure to 75 psia or 

above, the compression is often divided 
into two or more stages. Receivers 
are installed between the cylinders in 
which the successive compressions are 
effected in order to allow the com¬ 
pressed air to cool to atmospheric 
temperature before entering the suc¬ 
ceeding stage. If the compression 
could be carried out isothermally, no 
theoretical advantage would be gained 
by multistaging. When the compres¬ 
sion curve follows the ploy tropic 
PV n = C, the saving in work that results from two-stage compression is 
shown in Fig. 9:8. In the first stage, the air is compressed to an inter¬ 
mediate pressure Pi, the work per pound being, from Eq. (9:50), 



Fig. 9:8. Effect of multistaging in 
gas compression. 


w 


n 


1st stage 


71—1 


P\Vi 


1 - 


/pA (n_1)/n 


In the receiver between the first and second stages, the air is cooled at 
constant pressure P t - to atmospheric temperature so that the point i, from 
which the second stage of compression starts, lies on an isothermal 
through point 1, and PiVi = Ppq. The work performed on the air in the 
second stage, assuming n to have the same value as in the first stage, is 

1! 2d stage 7 P \V 1 

71—1 


> - (W 


The saving effected over single-stage compression is represented by the 
crosshatched area. The total work is the (negative) sum of the work 
areas for the first and the second stages, and this evidently becomes a 
minimum when the expression 















THE PERFECT GAS 


193 


has its minimum value. If we designate this function of Pi (Pi and P 2 
are constants since they are fixed as the limiting pressures of the total 
compression) by the symbol G, then the value of Pi obtained when 
dG/dPi equals zero will correspond to the minimum value of the function 
G and, therefore, to minimum total work. This mathematical operation 
shows that, for minimum total work, 


Pi 2 = P 1 P 2 


or 


Pi = P 2 = /P 2 V 
Pi Pi \Pj 


It will be noted that this ratio of pressures indicates that, for least total 
work, the work performed on the gas in the high- and low-pressure cylin¬ 
ders of a two-stage compressor will be equal. An analysis of three-stage 
compression will indicate that this principle of equal work performance, 
and therefore of equal pressure ratios for the individual stages, will again 
prevail, and thus 

P' = P" = P 2 = /P 2 V 
Pi P' P" \Pi) 

in which P' and P" are the absolute pressures at the end of the first and 
second stages, respectively. Similarly, for any number of compression 
stages X, the pressure ratio in each stage for minimum total work will 
become (P 2 /Pi) 1/X . Substituting (P 2 /Pi) 1/X for P 2 /Pi in Eq. (9:50), the 
work per stage may be calculated. But equal work is performed in each 
of X stages, and the total work of staged compression between the pres¬ 
sures Pi and P 2 is 

TTr XU J y 

* 1 X stages T P iF 1 

n — 1 


a 


(7i—1)/ Xn 


(9:53) 


Example 9:10. Calculate the power theoretically required for the two-stage com¬ 
pression of 100 lb of air per minute from 14 psia, 70°F, to 350 psia when n = 1.3. 
Compare with result obtained in part c of Example 9:9. 

Solution. Substituting RTi for Piv i in Eq. (9:53), 


Work per min 


MRTi 

n — 1 

24,500,000(1 


^p^in-D/Xn 

= [ (2) 0 (1 3 3 -] (100) (53.3) (530) [l - (^)° V2 ' 
— 1.45) = —11,000,000 ft-lb/min, or 333 hp 


For two-stage compression, the saving is 410 — 333 = 77 hp, or about 19 per cent. 


9:11. Clearance Factor and Volumetric Efficiency. Equations (9:49) 
to (9:51) state the theoretical work required for the compression of 1 lb 
of a gas in a cylinder without clearance. The introduction of clearance, 
necessary in the reciprocating compressor, will not increase the theoretical 
work of compression per pound of gas cleared through the discharge valve 
since that portion of the charge which is trapped in the clearance space 












194 


BASIC ENGINEERING THERMODYNAMICS 


requires work during compression but restores an equal amount during its 
eventual expansion to suction pressure. The effect of clearance is to 
increase the necessary piston displacement for a given capacity and 
therefore to increase somewhat the work required by the actual com¬ 
pressor, due to its irreversibilities. 

In Figure 9:9 is shown a theoretical indicator diagram of an ideal com¬ 
pressor with clearance. F 3 is the clearance volume, Fi — Vz is the piston 
displacement, and it is assumed that the value of n during expansion of the 

gas trapped in the clearance space is 
the same as during the compres¬ 
sion from Vi to F 2 . At point 4 the 
intake valve opens in response to 
lowered pressure within the cylinder, 
and the fresh charge mixes with the 
gas that was trapped in the clear¬ 
ance space. The assumption that n 
is the same for expansion 3-4 as for 
compression 1-2 is necessary if ir¬ 
reversibility is to be avoided during 
the process 4-1 due to the mixing of 
bodies of gas at different tempera¬ 
tures. 1 

The clearance is usually expressed, in terms of the piston displacement, 
as Vz = CV D , in which V D represents the piston displacement and is equal 
to Vi — Vz. The quotient (Fi — V A )/V D measures the ratio of the vol¬ 
ume of the fresh charge drawn into the cylinder per cycle, measured at 
suction pressure P i, to the piston displacement of the compressor and is 
called the clearance factor. It may be calculated in terms of the clearance, 
as follows: 


P 



pression. 


Vi = V D + cv D 


PzVz n 


P iVs 



Clearance factor = 




(9:54) 


1 Since, in the actual compressor, the temperature of the cooling fluid in the jacket 
continues to remain below that of the gas within the cylinder for the expansion 3-4 
as for the compression 1-2, the heat flow may be expected to remain negative, i.e., 
from the gas to the jacket, over at least much of the expansion. This would cor¬ 
respond to a value of n greater than k for the actual expansion of the trapped portion 
of the charge, instead of less than k as for the compression 1-2. Another somewhat 
unrealistic feature of the assumption is that the temperatures at 3 and 2 are assumed 
equal, and thus no cooling of the charge is assumed to take place during its delivery. 












THE PERFECT GAS 


195 


The term free air refers to the volume of air handled by an air com¬ 
pressor as measured at atmospheric pressure and temperature. This 
volume is less than as measured at suction pressure because of the differ¬ 
ential of pressure, P a — Pi, necessary to maintain flow from the atmos¬ 
phere into the cylinder during the suction stroke. The ratio of the vol¬ 
ume of the fresh charge of air drawn into the cylinder per cycle, measured 
at atmospheric pressure P a (on the free-air basis), to the piston displace¬ 
ment is called the volumetric efficiency. The volumetric efficiency is a 
ratio slightly smaller than the clearance factor. In order to find its value, 
it is necessary to calculate the free-air volume as a proportion of the vol¬ 
ume when measured at suction pressure. It has been shown in Art. 9:9 
that Ti = T a . Therefore P a V a = P 1 V 1 and V a /Vi = Pi/P a . Thus, if 
the volume drawn into the cylinder per cycle, as measured at suction 
pressure (the clearance factor), is multiplied by the ratio Pi/P a , the result 
will be the equivalent free-air volume, or 


Vv 


Pi Vx - V 4 _ Pi [\ , „ nfP*Y 

Pa Vd Pa L + VV 


(9:55) 


in which rj v = volumetric efficiency, a decimal fraction 
Pi = suction pressure (within cylinder) 

P a = pressure of atmosphere 

An analysis of Eqs. (9:54) and (9:55) will indicate that both the clearance 
factor and the volumetric efficiency will decrease with increasing clear¬ 
ances and increasing pressure ratios of compression. 

Compressors are often rated on the basis of their capacity in cubic feet 
of free air per minute. If the volume of free air to be handled per stroke 
is divided by the volumetric efficiency, the result will be the necessary 
displacement volume of the compressor. It will also be noted that P a V a 
may be substituted for PiVi in Eqs. (9:49) to (9:51) for the calculation of 
the theoretical work per pound of air compressed. 

Example 9:11. A two-stage double-acting air compressor, operating at 120 rpm, 
is to compress 200 ft 3 of free air per minute from 14 to 350 psia. The clearance at 
both ends of both cylinders may be assumed to be 3 per cent and the value of n during 
compression as 1.3. Atmospheric pressure is 14.7 psia. Calculate (a) the optimum 
pressure at discharge from the low-pressure stage, ( b ) the theoretical horsepower 
required for each stage and for the total compression, and (c) the necessary piston dis¬ 
placements of the low- and the high-pressure cylinders. 

Solution: 


(a) 

(b) 


Pi _ (Pz\ x/x = /350V 
Pi “ \Pi) V 14 / 

TTlow-pressure cylinder per mm 


= 5 or Pi = (14) (5) = 70 psia 

-(s!) (14-7)(144)(200)[l-(^) O V '- 3 ] 
= 1,835,000(1 - 1.45) = -825,000ft-lb/min, or 25hp 






196 


BASIC ENGINEERING THERMODYNAMICS 


The same horsepower will be required for the high-pressure stage, making a total of 
50 hp. This calculation takes no account of a drop in pressure between the two 
cylinders; we shall assume this to be 1 psi. As an approximation, this difference 
of pressure will be divided equally between the two cylinders, making the discharge 
pressure from the first stage 70.5 psia, the suction pressure for the high-pressure 
cylinder 69.5 psia. Making these corrections, 


IT low-pressure cylinder per mm — 1,835,000 


(W1 


= —830,000 ft-lb/min, or 25.1 hp 


A check of the power required by the high-pressure cylinder will show about the same 
value. This gives a total of 50.2 hp which may be considered to differ negligibly from 
the 50 hp calculated above since both these values must be adjusted upward to provide 
for the irreversibilities of the real compressor. 

(c) For the low-pressure cylinder, the volumetric efficiency is, from Eq. (9:55), 

‘ Vv = ( 1 P 7 ) f 1 + °'° 3 ” °- 03 (tt) 1/1-3 ] = 0 - 952 t 1 - 03 “ (0.03)(3.46)] = 0.882 


The displacement volume may be calculated by dividing the volume of free air per 
minute by the volumetric efficiency and the number of charges compressed per minute, 


V Dlow- 


pressure cylinder 


200 

(0.882) (2) (120) 


0.945 ft 3 


Allowing for the piston rod, this volume corresponds to a bore and stroke of approxi¬ 
mately 12 and 15 in., respectively. 

For the high-pressure cylinder, 


Vv 



1 + 0.03 - 0.03 



= 0.918 


When the air has been cooled to atmospheric temperature in the intercooler after 
discharge from the low-pressure cylinder, the volume per minute is 


PaVa _ (14.7) (200) 
Pi 70.5 


41.6 cfm 


and the piston displacement of the high-pressure cylinder, 


V Dhigh-pressure cylinder 


41.6 

(0.918)(2)(120) 


0.189 ft 3 


Assuming that the stroke is the same as that of the low-pressure cylinder (15 in.), 
this displacement volume corresponds to a bore of about 5.25 in. 


9:12. Steady-flow Expansion of Gases. In Fig. 9:5, it is indicated 
that the air leaving the compressor is to be the supply for an air engine. 
The development of power from compressed air is not, of course, its only 
use, but compressed air has certain advantages as a source of power since 
it may be transmitted over relatively long distances without excessive loss 
of energy. 

The steady-flow expansion of a gas may be carried out in an engine or a 
turbine; in either case the maximum work obtainable from the expansion 
is the same. A review of Art. 9:8 indicates that Eqs. (9:49) to (9:51) are 
just as suitable for application to the expansion of gases as to their com- 












THE PERFECT GAS 


197 


pression. For the expansion, the initial state is at the higher pressure, 
and P 2 /P 1 is less than unity. Examination of the equations shows that 
this will be reflected in a change of the sign of the work. Except for this 
difference in sign, when steady-flow compression and expansion take place 
reversibly between the same limiting pressures and volumes, the work of 
expansion should equal the work of compression if n has the same value 
for both processes. 

Example 9:12. Assume the air discharged from the adiabatic compression of 
Example 9:9 to enter and expand to 14 psia in a reversible adiabatic turbine. What 
horsepower is delivered by the turbine? 

Solution. At exit from the compressor (and entrance to the turbine), the tem¬ 
perature of the air is 530 (Vir) 0-4/1 4 = 1330°R. Applying Eq. (9:49) to the expansion 
in the turbine: 


W per min = ^ ^ MRTi 1 — ( p- \ 


1 - 


/ 14 \ 0.286 -1 
\350/ 


= (gjj) (100) (144) (1330) 

= 24,850,000(1 - 0.398) = +14,950,000 ft-lb/min, or 453 hp 
This is the same as the horsepower required for reversible adiabatic compression. 


When a reciprocating engine is used to deliver power as the result of the 
expansion of compressed air, the valves must be operated mechanically 
rather than in response to pressure differentials. The volume of the 
compressed-air charge admitted to the cylinder per cycle is thus adjusted 
independently of the pressure ratio of expansion, and the pressure reached 
by the charge at the end of the power stroke of the piston is usually above 
the final exhaust pressure (the pressure of the atmosphere). The theo¬ 
retical indicator diagram of the air engine will then show a vertical drop 
in pressure from this pressure to atmospheric pressure as the exhaust valve 
opens at the end of the power stroke; this represents a process during 
which the first portion of the charge to escape from the cylinder is 
(irreversibly) throttled to atmospheric pressure as the valve opens. 
Exhaust then proceeds at atmospheric pressure until the exhaust valve is 
mechanically closed. , 

Although the power output of a given engine can be increased by ad¬ 
mitting a greater volume of air per cycle, the amount of work realized per 
unitvolume of compressed air supplied is less than if the adiabatic expansion 
is completed to atmospheric pressure. The cycle of the reciprocating 
engine and the effect of engine clearance are discussed in greater detail in 
Chap. 13, though with particular reference to a more important type of 
reciprocating engine, the steam engine. 

Problems 

In the following list of problems , the specific heats of a perfect gas will be assumed constant 
unless the problem states otherwise. 





198 


BASIC ENGINEERING THERMODYNAMICS 


1. Show that what has been called an ideal gas in earlier chapters is recognizable 
as a perfect gas. What relation exists between C and R ? 

2. Identify each of the ideal gases W, X, Y, and Z with a perfect gas listed in 
Table 9:1. 

3. Describe the surface represented by the primary equation of state of a perfect 
gas. What is the trace of this surface on the plane T — 0? On the plane T = 100? 
On the plane T = 500? What effect on the surface results from an increase in the gas 
constant? 

4. Using absolute temperatures as ordinates and absolute pressures as abscissas, 
prepare a chart showing the variation of pressure and temperature of a perfect-gas 
system during changes of state which take place at constant volume. What is the 
pressure of the system at T =0? Interpret in terms of molecular activity. 

5. A new temperature scale is devised. The volume of a perfect-gas system at 0° 
on this scale and a pressure of 20 psia is 11 ft 3 . At the same pressure, but a tempera¬ 
ture of 100 deg on this temperature scale, the volume of the system is 14 ft 3 . Locate 
absolute zero on this temperature scale. 

6 . A perfect-gas system is confined in a closed rigid tank under a pressure of 
50 psia. Its temperature, as measured on a certain temperature scale, is 170 deg. 
The pressure decreases as heat is withdrawn from the system until it becomes 30 psia; 
the corresponding temperature is —20 deg. At what temperature is absolute zero 
located on this temperature scale? 

7. What is the volume of 3 lb of oxygen at a vacuum of 5 in. Hg and a temperature 
of 60°F? Barometer = 14.7 psia. 

8 . A tank contains 3 lb of air at a temperature of 100°F. The internal volume of 
the tank is 8 ft 3 . What pressure is recorded on a pressure gage attached to this tank? 
Barometric pressure is standard. 

9. A certain perfect gas has a density of 0.068 lb/ft 3 at so-called “standard condi¬ 
tions” (p = 14.7 psia, t = 32°F). What is the value of its gas constant? 

10. At a pressure of 75 psia, 3 lb of nitrogen has a volume of 7 ft 3 . What is its 
temperature? 

11. As a certain perfect-gas system weighing 3 lb increases in temperature from 
100 to 210°F during a reversible constant-pressure process, 15,000 ft-lb of work is 
performed. What is the molecular weight of the gas? 

12. A certain perfect gas has a molecular weight of 60. How much work would 
result from a reversible constant-pressure process during which 5 lb of this gas 
decreased 10°F in temperature? 

13. Chlorine (Cl) has an atomic weight of 35.5. What is the gas constant for 
methyl chloride (CH 3 C1) in the pressure-temperature range in which it acts as a 
perfect gas? 

14. At atmospheric pressure and temperature, nitric oxide (NO) may be treated 
as a perfect gas. What is its density at standard atmospheric pressure and 70°F? 

15. The monatomic gas argon has an atomic weight of 39.9. Calculate its density 
at standard atmospheric pressure and 70°F. 

16. A tank with a volume of 10 ft 3 contains air at 400 psia and 70°F. A second 
tank having a volume of 1 ft 3 contains air at 14.7 psia and 70°F. The two tanks are 
connected, and the pressure is allowed to equalize. What is the final pressure if the 
final temperature is 70°F? What weight of air has been exchanged? 

17. An observation balloon weighs 750 lb, including its load but not its gas contents. 
The balloon is filled with helium until it just floats in air at standard atmospheric 
pressure and a temperature of 70°F. What weight of helium is required? What 
volume? Assume equilibrium with the atmosphere. 

18. One-tenth mole of a perfect gas is confined within a tank at 14.7 psia and 32°F. 
What is the internal volume of the tank? 


THE PERFECT GAS 


199 


19. A tank contains 2.016 lb of hydrogen at 100 psia and 70°F. Sixteen pounds of 
oxygen is forced into the tank. If the temperature of the mixture is 70°F, what is 
the total pressure of the mixture? What are the partial pressures of the hydrogen 
and the oxygen? 

20. As the result of a negligible spark, the mixture of Prob. 19 burns completely to 
H 2 O. Assuming the products of the combustion to be a perfect gas, what would be 
the pressure within the tank when the original temperature is again reached? What 
weight of H 2 O is formed? What volume? According to the steam tables, are these 
values consistent? State your conclusions as to what must have happened to the 
water vapor. On this new basis, what is the pressure within the tank? What does 
the tank contain? 

21. A tank contains 12 lb of solid carbon having a negligible volume and 32 lb of 
oxygen at a pressure of 14.7 psia and 70°F. A negligible spark ignites the carbon, 
and it burns completely to CO 2 . After the products of combustion have returned 
to a temperature of 70°F, what is the pressure within the tank? 

22. Air is a mixture of approximately 23.5 parts by weight of oxygen and 76.5 parts 
of nitrogen. In air having a total pressure of 14.7 psia, what are the partial pressures 
of the oxygen and the nitrogen? Calculate the "apparent” gas constant for air on 
the basis of the composition as stated. 

23. Five pounds of air increases in temperature from 60 to 140°F. During the 
process neither the pressure nor the volume is constant. What is the change of 
internal energy of the system? 

24. Two pounds of air undergoes a process during which the heat flow is +50 Btu 
and 10,000 ft-lb of work is performed by the system. What is the change of tempera¬ 
ture of the air? 

25. An air system weighing 1 lb has a pressure of 50 psia and a volume of 4 ft 3 . 
As the result of a certain process, its pressure changes to 70 psia and its volume to 
3.2 ft 3 . What is the change of internal energy of the system? What is its change of 
enthalpy? Repeat solution, but based on a weight of 1.5 lb. Compare your answers 
as based on the two system weights. Develop an equation expressing the change of 
internal energy of a perfect-gas system in terms of the gas constant, the ratio k, and 
the initial and final pressures and total volumes of the system, assuming the specific 
heats to be constant. Write a similar expression for the change of enthalpy. 

26. The ratio k for methyl chloride (see Prob. 13) is 1.2. What are the changes 
of specific internal energy and specific enthalpy when the temperature decreases by 
30°F? 

27. Write an expression for the change of internal energy of a perfect-gas system in 
terms of its weight, its gas constant, the ratio of the specific heats, and the initial 
and final absolute temperatures. Write a similar expression for the change of 
enthalpy. 

28. Can the specific heats c v and c v of a perfect gas increase with temperature? If 
they do increase, does the difference between them necessarily remain constant? To 
what effect would you attribute the increase? Can c v and c v decrease with 
temperature? 

29. Calculate the values of c v and c v for argon (see Prob. 15) at atmospheric levels 
of temperature. As a perfect gas, what are the maximum values of c v and c p which are 
possible for argon? 

30. Calculate the values of c v and c v for nitric oxide (NO) at atmospheric tempera¬ 
ture. As a perfect gas, what maximum values of c v and c p are conceivable for this gas? 

31. As the result of a reversible polytropic process, the pressure of an air system 
changes from 200 to 20 psia while its volume increases from 1 to 9 ft 3 . What is the 
value of n? What is the external work? What is the change of internal energy? 
The heat flow? The change of enthalpy? 


200 


BASIC ENGINEERING THERMODYNAMICS 


32. The pressure of a 1-lb air system increases from 14.7 to 50 psia, its temperature 
from 00 to 230°F, during a reversible polytropic process. What is the value of n? 
What is the external work? What is the change of internal energy? The heat flow? 
The change of enthalpy? 

33. A closed system consists of 1 lb of air. Its initial pressure is 30 psia, and its 
initial volume is 7 ft 3 . If, during each of the following processes, its volume increases 
to 10 ft 3 , find the heat flow, the change of internal energy, the change of enthalpy, 
and the external work in each case: ( a) reversible constant pressure; ( b ) reversible con¬ 
stant temperature; (c) reversible adiabatic; ( d ) adiabatic expansion into an exhausted 
space; (e) reversible with n = — 1; (/) reversible constant internal energy; ( g ) reversi¬ 
ble constant enthalpy. 

34. During the reversible polytropic compression oi a perfect-gas system, the 
temperature rises and heat enters the system. Is the value of n between 0 and 1, 
between 1 and k, between k and oo, or does n have a negative value? 

35. During the reversible poly tropic compression of an air system, the heat flow is 
out of the system and is equal to one-third of the external work. What is the value 
of n? 

36. One pound of air has an initial pressure of 100 psia and an initial volume of 
2.5 ft 3 . At the end of a reversible process that is represented by a straight line on a 
Pv diagram, the pressure is 20 psia, and the volume 9 ft 3 . What is the change of 
internal energy during the process? What is the heat flow? Between what values 
does the polytropic specific heat c n vary during the process? 

37. During an irreversible process which follows the state path Pv n = const, with 
n constant, the pressure of an air system changes from 120 to 30 psia and its volume 
from 2 to 6 ft 3 . During the process the specific heat is constant, and the total heat 
flow for the process is —10 Btu. What is the value of n? What fraction of the maxi¬ 
mum work possible in following the same state path is lost because of the irreversi¬ 
bility of the process? 

38. Calculate the change of entropy of the system in Probs. 25, 32, 33(a) to ( g ), 
and 36. 

39. A Carnot engine uses air as the working substance, operating between source 
and refrigerator temperatures of 500 and 60°F, respectively. The pressure and 
volume at the beginning of the isothermal expansion are 300 psia and 4 ft 3 . During 
the isothermal expansion, the change of entropy is 0.16. Determine the pressures 
and volumes at the end of each process of the cycle. Calculate the work of each 
process and the net work of the cycle. Calculate the heat flow from the source, the 
heat rejected to the refrigerator, and the net heat flow of the cycle. What is the 
efficiency of the cycle? 

40. What horsepower is theoretically required for the isothermal compression of 
1000 ft 3 of air per minute from 14.5 to 50 psia in a single-stage compressor? What 
horsepower if the compression is adiabatic? If n = 1.3? 

41. Compare the horsepower necessary to compress 800 cfm from 14.7 to 45 psia 
in a single-stage compressor for oxygen and for nitrogen. Assume n is the same during 
the compression oi both gases. What are the relative weights which are compressed 
per minute if the temperature at the beginning of compression is the same? 

42. What horsepower is theoretically required for the compression of 30 lb of air 
per minute from 14 to 900 psia in a single-stage compressor, if n = 1.32? Atmospheric 
temperature is 70 °F. 

43. A centrifugal fan handles 5000 ft 3 of air per minute against a total head of 5 in. 
of water. Calculate the theoretical fan horsepower. 

44. The total head against which a fan operates is the sum of the static (pressure) 
head and the velocity head (the difference of the kinetic energies per pound at intake 


THE PERFECT GAS 


201 


to and discharge from the fan, expressed in terms of the distance through which this 
energy would lift a unit weight). A propeller type of fan handles 25,000 ft 3 of air at 
14.7 psia and 70°F per minute, increasing it from negligible velocity to a velocity of 
25 mph. The static pressures at intake to and discharge from this fan, at the points 
where the velocities are measured, are balanced. Calculate the theoretical fan 
horsepower. 

45. Calculate the horsepower theoretically required as based on the data of Prob. 
42 if three stages of compression are used. What are the pressures at discharge from 
each stage? 

46. Prove that if G (see Art. 9:10) is to have a minimum value then Pi 2 = PiP 2 . 

47. What horsepower is theoretically required for the compression of 1000 ft 3 of air 
per minute from 14 to 200 psia with n = 1.3 if two stages of compression are used? 
What is the pressure at discharge from each stage? What is the saving over single- 
stage compression? 

48. A compressor with 3 per cent clearance operates at a suction pressure of 14.2 
psia when atmospheric pressure is 14.7 psia. The discharge pressure is 60 psia, and 
n = 1.32. What is the clearance factor? The volumetric efficiency? If this com¬ 
pressor handles 1400 ft 3 of free air per minute, is double-acting, and operates at 150 
rpm, what is its piston displacement? 

49. The total cylinder volume of a double-acting compressor with equal clearances 
at the two ends of the cylinder is 5.3 ft 3 . Its piston displacement is 5 ft 3 , and it makes 
80 rpm. The suction pressure is 14.3 psia, discharge pressure is 45 psia, atmospheric 
pressure is standard, and;?! = 1.3 for the compression. What is its capacity in cubic 
feet of free air per minute? What is its theoretical horsepower? 

50. One hundred cubic feet of air at 150 psia and 600°F is supplied per minute to a 
reversible adiabatic engine which exhausts at 14.7 psia. What horsepower does the 
engine deliver? What is the temperature of its exhaust? 

51. If air at the state and at the rate described in Prob. 50 is delivered to an engine 
which has a cylinder surrounded by a water jacket containing water at 70°F, would 
you expect to find the value of n increased or decreased in comparison with its value 
for the adiabatic engine? Would the power output of the engine have been increased 
or decreased? Assuming that lubrication difficulties could be overcome without 
water-jacketing, would you consider it desirable to water-jacket an engine cylinder? 

Symbols 

c n polytropic specific heat, general 
c p specific heat at constant pressure 
c v specific heat at constant volume 

C a constant; also, ratio of clearance volume to piston displacement 
/ impact force of a molecule «. 

F impact force of a collection of molecules 
h enthalpy of unit mass 
H enthalpy of a system 
J proportionality factor 
k ratio of the specific heats 
m molecular weight; mass of a molecule 
M mass of a system; mass rate of flow 
n a constant 

N number of molecules in a system of unit volume 
p pressure, psi 

P pressure, psf; pressure in general 
Q heat flow; rate of heat flow 


202 


BASIC ENGINEERING THERMODYNAMICS 


R gas constant 
s entropy of unit mass 
S entropy of a system 
T absolute temperature 
u internal energy of unit mass 
U internal energy of a system 
v specific volume 

V volume of a system; also, volume rate of flow 
Vd piston displacement volume 

V velocity 

W work; rate of work delivery 
X number of stages of compression 
z elevation 

Greek Letters 

rj v volumetric efficiency 
n Joule coefficient 
hj Joule-Thomson coefficient 
p mass density 

Subscripts 

a atmospheric; free-air basis 
i intermediate 
m molar; of a mole 
n polytropic (see c„) 
p constant pressure 
rev reversible 
T constant temperature 
u constant internal energy 
v constant volume 


CHAPTER 10 


MIXTURES OF GASES AND VAPORS 
MIXTURES OF PERFECT GASES 

10:1. Basic Principles. The engineer often deals with mixtures of 
gases, and it is necessary to develop methods by which the properties of 
such mixtures may be determined. Although air is perhaps the outstand¬ 
ing example of these mixtures, many others may be cited, including air- 
fuel mixtures and the products of combustion. When the pressures of the 
various constituents are uniformly low as compared with their critical 
pressures and when they remain in the gaseous phase throughout the 
engineering process, they may be treated as perfect gases with the accu¬ 
racy usually sufficient for the engineering calculation. It has been indi¬ 
cated in Chap. 9 that the thermodynamic method of handling these mix¬ 
tures consists in finding an average molecular weight (the weight of 1 
mole) of the mixture and thereafter treating the mixture as if it were com¬ 
posed of a single gas having molecules of a relative weight equal to this 
average; the mixture thus becomes, in effect, a new “pure substance.” 

According to the principle first stated by Dalton, any gas, in a mixture 
of gases, acts as if it were in a vacuum and the other gases were not 
present. The basis of this principle, as it applies to the perfect gas, was 
outlined in Art. 9:2. Gibbs restated the Dalton principle in the more 
specific and usable form which we know as the Gibbs-Dalton law: 

The pressure, internal energy, and entropy of a mixture of gases are, 
respectively, equal to the sums of the pressures, internal energies, and 
entropies of the component gases when each occupies alone the volume of the 
mixture at the temperature of the mixture. 

Although the Gibbs-Dalton law is based on the concept of the perfect 
gas, experience has shown that it holds for real gases with close accuracy, 
especially at low pressures. 

Let us consider a simple system consisting of a mixture of a number of 
gases which we shall designate as 1, 2, 3, etc. These gases occupy a com¬ 
mon volume, the volume of the mixture, and we may accordingly write 

V l = V 2 = V 3 = • • • = V n = V m (10:1) 

in which the subscripts 1 to n refer to the component gases and m to the 
mixture. Assuming the mixture to be in equilibrium, the temperature of 

203 


204 


BASIC ENGINEERING THERMODYNAMICS 


all components must also be the same and equal to the temperature of the 
mixture, or 

Ti = T 2 = Tz = • • • = T n = T m (10:2) 

According to the Gibbs-Dalton law, 

Pi + P 2 + Pz + • • • +Pn = Pm (10:3) 

in which Pi to P n are the partial pressures created by the bombardment of 
the molecules of the respective gases and P m is their sum, the total pressure 
of the mixture. Further, from the same law, 

Pi + U 2 + Uz +••*+. U n = U m (10:4) 

and 

-b S 2 -j- Sz H - * * ■ -b S n = S m (10:5) 

Also, Hr = Pi + PiFi/J = Pi + PiVJJ, H 2 = P 2 + PYVJJ , etc., 
and since the sum of the internal energies of the components is U m and the 
sum of their partial pressures is P m , 

Hi + H 2 + Hz -b • • • -b H n = U m -f- — -j — = H m (10:6) 

Finally, the mass of the mixture is equal to the sum of the masses of its 

components, 

or 

M\ + M 2 + M 3 + * * ■ -b M n = M m (10:7) 

The relations expressed by Eqs. (10:1) to (10:7) are the foundation under¬ 
lying the calculation of the properties of the mixture. 

10:2. Definitions. A mole of any gas (or, indeed, of any liquid or solid) 
consists of a standard number of molecules (see Art. 9 : 2 ). This is true 
whether the gas is composed of molecules which all have the same mass or 
whether, as in a mixture, it is a collection of molecules of different mass. 
If the mixture is composed of perfect gases, the volume of the mole will 
be the same at the same pressure and temperature for any mixture as for 
the component gases, considered individually. This characteristic of the 
mole makes it a convenient unit on which to base definitions of mixture 
composition and calculations of mixture properties. 

In describing the composition of a mixture of gases, various methods 
may be used. For instance, the amounts of the various components may 
be stated in terms of their mole fractions. The mole fraction of a compo¬ 
nent gas in a mixture of gases is the ratio of the number of moles of that gas in 
a given volume of the mixture to the total moles of the mixture in that volume. 
The analysis of a mixture to obtain the mole fractions of its component 
gases can be readily made, for, at the same pressure and temperature, 
equal volumes of all perfect gases (or mixtures of perfect gases) contain 



MIXTURES OF GASES AND VAPORS 


205 


the same number of molecules and therefore the same number of moles. 
In practice, these analyses are made by means such as the Orsat analyzer. 
The method used is to separate a sample of the mixture and measure its 
volume at atmospheric pressure and temperature. One of the component 
gases is then chemically absorbed from the mixture, and the remaining 
volume is measured at the same pressure and temperature; the difference 
between this and the original volume measures the amount of that compo¬ 
nent in the mixture, the ratio of the difference to the original volume of 
the mixture being the mole fraction Xi assigned to that component. The 
next component is then absorbed from the remainder of the sample, and 
x 2 , its mole fraction, is obtained by comparing the decrease in volume due 
to its removal with the original volume of the mixture sample. This is 
continued until account is taken of all components of the mixture. 

The result of an Orsat analysis is often called the volumetric analysis , 
or proportion by volume , but corresponds to a description of the mixture in 
terms of its mole fractions. Because each of the component gases 
actually occupies the full volume of the mixture, the use of “mole frac¬ 
tion’’ is to be preferred to “proportion by volume.” 

Another method of describing gas mixtures is by means of the gravi¬ 
metric analysis, or proportion by weight, of the various constituents. 
Referring to Eq. (10:7) and the notation used in that equation, the pro¬ 
portion by weight of component 1 is, for example, M\/M m . 

10:3. Calculation of Mixture Properties. The state of a mixture of 
gases is usually defined in terms of its temperature T m and its pressure P m . 
The composition of the mixture may be presented either in terms of the 
mole fractions of its component gases (their proportions by volume) or, 
less frequently, in the form of a gravimetric analysis. Assuming that 
their mole fractions have been determined, the weights of the components 
in 1 mole of the mixture are, respectively, Xi m x , x 2 m 2 , x^m h etc., where 
m h m 2 , m 3 , etc., are the respective molecular weights (the weights of 1 
mole) of the component gases. But 


xpux -j- x 2 m 2 -f x 3 ra 3 + • • • + x n m n = m m (10:8) 

4 

in which m m is the weight of 1 mole of the mixture (the “apparent ” molec¬ 
ular weight). When m m has been obtained in this manner, the products 
oqrai, x 2 m 2 , £ 3 m 3 , etc., may be compared with m m to obtain the gravimetric 
analysis, as follows: 


Mi _ Ximi 

AI m m m 


M 2 _ x 2 m 2 
M m m m 


etc. 


(10:9) 


If, on the other hand, the composition of the mixture is described in 
terms of a gravimetric analysis, the corresponding division into mole trac¬ 
tions may be obtained since the proportion by weight from the gravi- 





206 


BASIC ENGINEERING THERMODYNAMICS 


metric analysis is the weight of a component gas in 1 lb of the mixture and 
the division of this weight by the molecular weight of the component gas 
gives the number of moles of that gas in 1 lb of the mixture. The sum of 
these quotients for all of the components is therefore the number of moles 
of the mixture in 1 lb of the mixture, or 1 /m w . Thus, 


M i 1 ^ M 2 1 M% 1 | | M n 1 _ 1 

M m mi M m m 2 M m m 3 M m m n m m 


( 10 : 10 ) 


from which m m may be calculated. Then, from Eq. (10:9), 


Mi m rn 
Xl = T7- 

M m mi 


M 2 rn m 

M m m 2 


etc. 


( 10 : 11 ) 


It has been shown above how m m , the weight of 1 mole of the mixture, 
may be obtained from the proportions of the mixture, whether those pro¬ 
portions be expressed in terms of mole fractions or in the form of a gravi¬ 
metric analysis. Dividing this apparent molecular weight into the uni¬ 
versal gas constant gives the value of R, the gas constant that applies to 
the mixture. Thereafter the equations of Chap. 9 will apply, exactly as 
if the mixture had been composed of a single kind of molecule. The 
values of c v and c v will, for the mixture, be the weighted average of their 
values for the component gases, as follows: 


c 


Pm 


c v „ 


Mi Mt , M z M„ 

M m Cp ' + M m Cv ' + M m Cpi + ' ' ' + M m Cp " 
Mi Mi M t , ,M n 

M m C *■ + M m + M m C ’> + ' ' ' + M m C *’” 


( 10 : 12 ) 

(10:13) 


According to the Dalton principle, PiVi = MiRiT h P 2 V 2 = M 2 R 2 T 2 , 
etc., and, substituting V m for V h V 2 , etc., T m for T i, T 2 , etc., and 1545 /m 
for R, 


P 


154:5MiTm 

mi 


PPYm 


154 bM 2 T m 
m 2 


etc. 


Also PmVm — 154 5MmT m /m m , and, dividing, 

Pi = Mi rrim _ P 2 
Pm M m mi Xl P m 


etc. 


(10:14) 


Pi, P 2 , etc., are the partial pressures of the component gases, and Eq. 
(10:14) shows that these can be obtained by multiplying the total pressure 
of the mixture by the respective mole fractions of its components. 

Example 10:3. An Orsat analysis of the products of combustion resulting from the 
burning of a fuel gas yields the following percentages: C0 2 — 10; 0 2 — 7; N 2 — 83. The 
water and water vapor have been removed from the sample before this analysis was 
made, and the analysis therefore presents the composition of the dry products, (a) 
Calculate the gravimetric analysis, the apparent molecular weight, and the gas constant 

















MIXTURES OF GASES AND VAPORS 


207 


for the dry mixture. (6) Assuming the total pressure of the dry mixture to be 14.7 
psia and the temperature 100°F, find the partial pressure of each component and the 
specific volume of the mixture, (c) Calculate the specific heats, c p and c v , of this 
dry mixture. 

Solution: 

(a) The Orsat analysis is a volumetric analysis, and the corresponding mole fractions 
are obtained by dividing the percentages by 100. The following tabular form of 
solution will be found convenient : 


No. 

Component 

Mole 

fraction 

X 

Mol. wt. 

rn 

Wt. per mole 
of mixture 

xrn 

Gravimetric 

analysis 

M/M m 

1 

C0 2 

0.10 

44 

4.40 

0.147 

2 

o 2 

0.07 

32 

2.24 

0.075 

3 

N 2 

0.83 

28.016 

23.25 

0.778 

Total. 

. . . 



29.89 

1.000 


Rr 


1545 

'Wlm 


1545 

29.89 


51.7 


( 6 ) pi = Xip m = (0.10) (14.7) = 1.47 psia 

P 2 = X 2 p m = (0.07) (14.7) = 1.03 psia 

p 3 = xzp m = (0.83) (14.7) = 12.20 psia 

14.7 psia 


R m T m = (51.7) (560) 
P m (14.7) (144) 


13.65 ft 3 /lb 


(c) The specific heats of the component gases may be taken from Table 9:1. Then 


c Pm = (0.147) (0.201) + (0.075) (0.217) + (0.778) (0.248) = 0.239 
c„ m = (0.147) (0.156) + (0.075) (0.155) + (0.778) (0.177) = 0.172 

The specific heats of air are often used to apply to the dry products of combustion 
in approximate calculations. A comparison of the values calculated above with the 
corresponding values for air indicates why this is possible. Although the dry products 
vary somewhat in composition as various fuels are burned with different weights of 
air supply per pound of fuel, the variation of c p and c v , within the usual limits of 
composition, is minor. 

10:4. Adiabatic Mixing Processes for Perfect Gases. To aid our 

study of the nonflow mixing process, let us refer to Fig. 10:1, which shows 
a rigid insulated container that has been divided into a number of com¬ 
partments (three are shown in the figure, but the number need not be 
limited) by rigid insulated partitions which are removable and occupy a 
negligible portion of the total volume of the vessel. Originally, gas 1 is 
confined in compartment a at pressure P a and temperature T a , gas 2 in 
compartment b at Pb and Tb, etc. 

For the first example, we shall assume that the individual pressures and 
temperatures of all these gases are the same. If the partitions are now 
























208 


BASIC ENGINEERING THERMODYNAMICS 


removed, each gas will expand to occupy the total volume of the container. 
The resultant mixing process is adiabatic since heat flow is prevented by 
the insulation that covers the vessel walls; these walls are rigid, and no 

work is exchanged with any ex¬ 
ternal system. Therefore the in¬ 
ternal energy of the resulting mix¬ 
ture is the sum of the original 
internal energies of its parts before 
mixture took place. But these are 
perfect gases, and their internal 
energies are accordingly functions 
of their temperature alone and 
independent of the volume they occupy. Therefore the temperature of 
the mixture must be identical with the original common temperature of 
its components. 

Each gas has expanded at constant temperature, and Boyle’s law will 
apply. Designating the original equal pressures P a , Pb, etc., as P and the 
partial pressures of the components in the final mixture as Pi, P 2 , etc., we 
may write 

P«F„ = PV a = PiV m or = — PVb = P 2 V m or ^ = A 

But since the original pressures and temperatures were equal, V a /V m = Xi, 
Vb/Vm = x 2 , etc. Therefore Pi/P = Xi, P 2 /P = x 2 , etc. From Eq. 
(10:14), P\/P m = Xi, P 2 /P m = etc., and P m must therefore equal P. 

The mixing process has been adiabatic, but an analysis will indicate that 
it is irreversible, since external systems, not affected during the mixing 
process, must be called upon to supply energy if each of the component 
gases of the mixture is again to be restored to its individual compartment. 
We should therefore expect to find that the entropy had increased during 
mixing and was greater than the sum of the original entropies of the parts. 
Although the temperature of these parts has not changed during mixing, 
their pressures have individually decreased and their volumes have 
increased. A convenient choice may be made from Eqs. (9:44a) to 
(9:47a) for the calculation of the increase of entropy of each component; 
these individual increases may be added to obtain the total increase of 
entropy due to mixing. 

Let us now consider a more complicated mixing process in which neither 
the pressures P 0 , Pb, etc., nor the temperatures T a , T b , etc., are equal, 
though their magnitudes are individually known, as are the volumes V a , 
V b , etc., and the kind of gas confined in each compartment. The weight 


'//////////// 


'////////////////////////; 

!• 

IWW 

b 

'I 


Fig. 10:1. Adiabatic nonflow mixing. 










MIXTURES OF GASES AND VAPORS 


209 


of gas in each compartment may be calculated and a gravimetric analysis 
of the mixture that will result when the partitions are removed thus 
obtained. Applying methods suggested in the preceding article, the gas 
constant of the mixture may be computed. The internal energy of the 
mixture is again the sum of the original internal energies of its parts, and 
though, in this case, each gas changes in internal energy, the sum of these 
changes must therefore be zero. Writing this relation in the form of an 
equation, 


MiC vl (T m - Ta) + M 2 c V2 (Tm ~ T b ) + • • • + M n c Vn (T m - T n ) = 0 

(10:15) 

in which T m , the final temperature of the mixture, is the only unknown 
and therefore may be directly computed. This leaves P m as the only 
unknown in the equation P m Vm — M m R m Tm and makes its calculation 
possible. The increase of entropy for this adiabatic mixing process must 
not only account for the irreversibility of the mixing alone but also reflect 
the fact that the system was not originally in temperature and pressure 
equilibrium. 

Example 10:4. The three compartments into which a closed insulated container 
is divided contain, respectively, 1 lb of hydrogen at 200 psia and 60°F, 1 lb of helium 
at 100 psia and 100°F, and 1 lb of carbon dioxide at 15 psia and 120°F. The parti¬ 
tions are removed, and a mixture is formed, (a) What are the mole fractions of each 
component in the mixture and the apparent molecular weight and gas constant of the 
mixture? ( b) What is the temperature of the mixture? ( c ) Find the partial pres¬ 
sures of each component of the mixture. ( d ) Calculate the specific heats of the 
mixture and their ratio, k. ( e ) What change of entropy results from mixing? 

Solution: 

(a) A tabular form of solution is again convenient: 


No. 

Component 

Gravimetric 

analysis 

M/M m 

Mol. wt. 

m 

Moles per lb 
of mixture 
M/M m ni 

Mole 

fraction 

1 

Ho 

0.333 

2.016 

0.1655 

0.645 

2 

He 

0.333 

4.002 

0.0833 

0.325 

3 

C0 2 

0.333 

44 

0.0076 

0.030 

Total . 




0.2564 

1.000 






m m 


1 

0.2564 


= 3.90; R m 


1545 

3.90 


396 


(b) Applying Eq. (10:15), 


(1) (2.43) (t m - 60) + (1)(0.755)(2m - 100) + (1) (0.156) (« m - 120) = 0 






















210 


BASIC ENGINEERING THERMODYNAMICS 


Solving, 
t m = 72°F 

, ^ rr | t/ - i ~\t MaRaTa , MbRbTb , M C R C T c 

(c) F m = Fa + Kb + F c = - 5 -b - n -1-5- 

la i b 1 c 

(1) (766.6) (520) (1) (386) (560) (1) (35.1) (580) = 

(200) (144) ^ (100) (144) ^ (15) (144) 


Pm — 


M m R m T m 


= 115 psia 


(3) (396) (532) 

144 144F m (144) (38.27) 

pi = Xip m = (0.645) (115) = 74.0 psia 

Pi = x 2 p m = (0.325) (115) = 37.5 psia 

pi = x-.\p m = (0.030) (115) = 3.5 psia 

115 psia 

0 d) Cp m = (0.333)(3.42) + (0.333)(1.25) + (0.333)(0.201) = 1.624 
c Vm = (0.333) (2.43) + (0.333) (0.755) + (0.333) (0.156) = 1.114 


= — = Fri - lA6 

Cv m 1.114 


(e) Applying Eq. (9:47a), 

T m 


A $1 = Mc pi loge 


T 

-L a 


MRi , P i /i\/o ,io\ i 532 
— log.-p- = (1)(3.42) log. m 


(1) (766.6) 74.0 


778 


= +0.078 + 0.978 
AS 2 = (1)(1.25) log e |E - 

A1S3 = (1) (0.201) log e |h 


(1)(386). 37.5 nncn . n/IOK 

- -■ ?78 - log«+QQ = -0-059 + 0.485 

(1X35.1+ 3.5 


778 


log e T+ = -0.017 + 0.066 
15 


200 
= +1.056 

= +0.426 
= +0.049 


AS = +1.531 


The foregoing analysis is concerned with the adiabatic nonflow mixing 
process. The mixing of streams of gases in steady flow is more charac¬ 



teristic of the usual engineering situation and is illustrated in Fig. 10:2. 
Section i is a section across all of the incoming streams of gas; section o 
intersects all outgoing streams. Applying the steady-flow energy equa¬ 
tion [Eq. (3:5)], we may write 









































MIXTURES OF GASES AND VAPORS 


211 


X M ( h + Si + 

* 0 

in which ^ represents the summation for all incoming streams and ^ the 

1 0 
corresponding summation for those leaving the mixing chamber. 

If the mixing is adiabatic and no external work is performed, the Q and 
W terms in Eq. (10:16) will be zero. In addition, the difference between 
streams in stored energy due to motion and elevation will often be 
negligible in comparison with differences of enthalpy, and, for this set of 
circumstances, Eq. (10:16) simplifies to 

^ Mh = Y Mh (10:17) 

i o 

It will be observed that the application of Eqs. (10:16) and (10:17) is not 
restricted to perfect gases but may be extended in application to the 
steady-flow mixing of any combination of fluids, including liquids. 

10:5. The isentropic process for a mixture of gases having separately 
different values of k, the ratio of their specific heats, is of special interest. 
When, for example, the mixture is composed of a monatomic and a 
diatomic gas, the value of k for the mixture will be intermediate between 
1.67 and 1.4. The isentropic for the mixture will follow the PV relation 
PV k = C, where k has this intermediate value. But ki > k > k 2 , where 
the subscript 1 refers to the monatomic and 2 to the diatomic component. 
In other words, although the process is isentropic for the mixture, it is not 
isentropic as regards the separate components. We may write 

A S m = 0 = ASi T AS 2 or A*S»i = — AS 2 (10:18) 

A$i and A*8 2 may be calculated by applying the appropriate equation from 
Table 9:2, substituting k for n and k\ and k 2 , successively, for k in the 
equation for the change of entropy during a polytropic process. 

Example 10:5. The mixture of Example 10:4 expands isentropically to a final 
pressure of 20 psia. Find the individual entropy changes of its components. 

Solution. For each component, n = k m = 1.46 for this expansion. The tem¬ 
perature at the end of the expansion is T 2 = 7 7 i(P 2 /Ti) ri_1/ ' 1 = 532 (tts) 0 ' 4671,46 == 
(532) (0.55) = 293°R. From Table 9:2, 

ASh, = Me, log. = (1) (2.43) ( ^ "J, 41 ) log. 0.55 

= (0.264) (-0.597) = -0.158 

ASh. = (1) (0.755) ( ^kTi" ) l0g * °- 55 = (-°- 328 )(-°- 597 ) = +0.196 

AiScoi = (1)(0.156) Pflg-Jf 9 ) l°ge0.55 = (0.058)(-0.597) = -0.035 

A S m = +0.003 

The discrepancy in AS m (AS m should total zero) is due to slide-rule error. 


X- 


2 Jg ' J 


) 


h -f- + -j 1 + iWo (10:16) 









212 


BASIC ENGINEERING THERMODYNAMICS 


MIXTURES OF A GAS AND A VAPOR 

10:6. Characteristics of Gas-vapor Mixtures. According to the dis¬ 
tinction that has been made between the meanings of the terms gas and 
vapor, as we shall use them (see Art. 9:1), the vapor component of a gas- 
vapor mixture might be expected, in the course of an ordinary thermo¬ 
dynamic process, to undergo a process of condensation or to increase its 
proportion in the mixture owing to evaporation. At least a part of that 
component would then be, occasionally, in the liquid phase. The mix¬ 
ture of gases is a pure substance, but gas-vapor mixtures cannot be so 
treated since the heavier liquid phase of the vapor will tend to settle, and 
the mixture will not be chemically homogeneous, as is required for classifi¬ 
cation as a pure substance. Examples of gas-vapor mixtures of engineer¬ 
ing importance are atmospheric air, which is a mixture of the dry air of 
Chap. 9 with water vapor, the mixtures of fuel and air which are produced 
in the operation of the liquid-fuel carburetor, and the products of com¬ 
bustion resulting from the burning of a hydrogen-bearing fuel. Some¬ 
times, for the purpose of more effective cooling, water is injected directly 
into the cylinder of an air compressor or an internal-combustion engine; 
its evaporation produces a gas-vapor mixture. Of these, the mixture of 
most general engineering interest is the air of the atmosphere, and a num¬ 
ber of the special thermodynamic terms that are applied to gas-vapor 
mixtures in general are especially descriptive of atmospheric air. Because 
of this and since, in addition, atmospheric air is the gas-vapor mixture 
most familiar to the reader from the standpoint of personal observation of 
its behavior during changes in state, the discussion to follow is related 
directly to the moist-air mixture; the principles that develop will apply, 
in general, to many other gas-vapor mixtures. 

Let a pan of water be placed in a room. A separation surface will be 
maintained between the water and the air above because of the liquid’s 
property of surface tension. It will normally be observed that the 
amount of water in the pan gradually decreases, and it is evident that it 
must have escaped in the form of a vapor; yet no boiling of the liquid has 
taken place. The explanation is that the separation surface is not an 
impervious wall and therefore permits the passage of occasional molecules 
from among the immense number that strike it. This surface is bom¬ 
barded from both sides—from below, by water molecules and, from above, 
by both the molecules of the air and those of the water vapor already 
present in that space. The air molecules cannot condense to form a 
liquid and therefore cannot pass the wall; the water-vapor molecules may 
occasionally pass and, as a result, become liquid molecules below the 
surface. At the same time, of course, liquid molecules from below the 
surface are escaping into the air above and so becoming vapor molecules. 


MIXTURES OF GASES AND VAPORS 


213 


If their rate ot passage exceeds the rate at which vapor molecules assume 
the liquid phase, evaporation takes place; if the opposite is the case, 
condensation occurs, and the proportion of vapor molecules in the space 
above will decrease. These rates will depend upon the pressure exerted 
against the separation surface by the bombardment of the liquid and of 
the vapor molecules, respectively; when these pressures are balanced, the 
number of molecules escaping from the liquid will equal the number 
returning to the liquid and neither evaporation nor condensation will 
be observable, though an exchange of molecules will still be taking 
place. 

The pressure exerted against the separation surface by the liquid mole¬ 
cules will depend upon their temperature; the pressure created by the 
vapor molecules is a function of both their temperature (assumed to be 
the same as the temperature of the liquid) and their concentration, 
increasing as that concentration increases. These pressures are balanced 
when the concentration of vapor molecules is such that their pressure is 
the saturation pressure corresponding to their temperature and the tem¬ 
perature of the liquid below the surface, according to the pressure-tem¬ 
perature relation (see Chap. 7) of the substance. For example, for water 
vapor at 70°F, this pressure is 0.3631 psia. This saturation pressure 
forms a part of the total pressure of the mixture of air and water vapor and 
is, of course, the partial pressure of the water vapor in that mixture. It 
is also the highest pressure possible for water vapor at the given tempera¬ 
ture and therefore the highest partial pressure possible for the water vapor 
in an equilibrium mixture at that temperature. For if the vapor pressure 
were higher, condensation would take place until a state of equilibrium 
was attained. Thus the concentration of vapor molecules that is associ¬ 
ated with the maximum vapor pressure is the maximum concentration 
that is possible in an air-water vapor mixture that is in equilibrium, and 
the air is said to be saturated . Note also that this concentration is the 
same as that which is characteristic of “saturated” vapor, in the sense in 
which that term was used in Chap. 7; the properties of the vapor portion 
of the mixture may therefore be obtained from tables of the properties of 
steam. 

A summary of the conclusions reached in the preceding paragraph, as 
they apply to the general gas-vapor mixture which is in equilibrium with 
the liquid phase of its vapor component, leads to an extension of the 
Gibbs-Dalton law to include the following statement: When a mixture of a 
gas and a vapor is in equilibrium with the liquid phase of the vapor component 
of the mixture, the density of the vapor is the same as the density of the pure 
vapor when in equilibrium with its liquid at the same temperature. 

Designating the vapor component as 1 and the gas as 2, Eqs. (10:1) to 
(10:7) may be applied to the saturated gas-vapor mixture, as follows: 


214 


BASIC ENGINEERING THERMODYNAMICS 



V m V gi V 2 
Vm = Vox + P2 

Um = Iiw, + M 2 w 2 (10:19) 

H m = M\h gi + M<Ji2 
S m = Mis gi + M 2 s 2 

If component 2 is a mixture of gases instead of a single gas, the properties 
of that mixture may be assembled, by applying the methods of pages 205 
to 207 of this chapter, before the properties of the gas-vapor mixture are 
investigated and this component can thereafter be treated as a single gas. 

Example 10:6. A saturated atmospheric-air mixture has a total pressure of 14.7 psia 
and temperature of 70°F. Calculate the partial pressures of the water vapor and the 
dry air and the weight of water vapor per pound of dry air in the mixture. Find the 
volume, internal energy, enthalpy, and entropy of the mixture per pound of its dry-air 
content. Assume a reference level, corresponding to zero enthalpy and entropy, of 
14.7 psia and 0°F for the dry-air portion, and, for the water vapor, use steam-table 
values. 


Solution. Partial pressure of the water vapor = p 0{ = saturation pressure at 
70°F = 0.3631 psia. Partial pressure of dry air = p 2 = p m — p g i = 14.7 — 0.3631 

= 14.34 psia. V m = v 2 = — p — = ^4 34 ) ( 144 ) = 13.65 ft 3 per pound of dry air. 
v 0i = 867.9 ft 3 /lb [Steam tables at 70°F] 

y | ^ 0 ^ 

Miv 0l = V m or Ml = —- = ‘ = 0.0157 lb water vapor per pound of dry air 

v 0l oo/.y 

H m = Mih 0l + M 2 h 2 = (0.0157) (1092.3) + (1) (0.24) (70 - 0) = 34 Btu 

At 14.7 psia and 0°F, the internal energy of 1 lb of dry air = h — ^ = 0 — ~ - 

J J 


-(53.3) (460) 
778 


= —31.4 Btu. Then 


Um = Miu 0i + M 2 u 2 = (0.0157)(1034.1) + 1[ —31.4 + 0.170(70 - 0)] = -3.2 Btu 
Checking, 

Um . H „ _ P=y. . 34 - Q4 : 7) (144) (13.65) . 34 _ 37 . 2 . _ 3 . 2 Btu 


J 

T R 


778 


s 2 = c p log e -- -J loge y = 0.24 log, 


530 53.3 


log, 


14.34 


= 0.0338 + 0.0017 


0 ^ r 0 460 778 14.7 

= 0.0355 

Sm = Mis gx 4 - M 2 S2 = (0.0157) (2.0647) + (1) (0.0355) = 0.0324 + 0.0355 = 0.0679 


When the concentration of vapor molecules is less than the maximum, 
the vapor pressure is below the saturation pressure corresponding to the 
temperature of the mixture and the vapor component is therefore in the 
form of a superheated vapor. In Art. 7:9, it was pointed out that, at low 
pressures, lines of constant temperature on the Mollier chart become 












MIXTURES OF GASES AND VAPORS 


215 


horizontal, or constant-enthalpy, lines. This means that the enthalpy of 
the water vapor, like that of the perfect gas, is, for these pressures, a func¬ 
tion of the temperature alone and suggests that, whether saturated or 
superheated, the water-vapor component of the atmospheric-air mixture 
may be treated as a perfect gas with negligible error. Thus the data of 
the steam tables may be extended to superheat states not shown in the 
table. The lowest pressure tabulated in Table 3 of the Keenan and Keyes 
tables is 1 psia, but the pressure of the superheated steam in the atmos¬ 
pheric-air mixture is usually much less than this pressure. From the 
above, the specific enthalpy (or the internal energy) of this superheated 
steam is the same as foi; saturated steam at the same temperature, and h g 
at the temperature of the mixture may be substituted. As an example, 
suppose that an atmospheric-air mixture (not saturated) has a tempera¬ 
ture of 80°F. Its water-vapor content is superheated steam, but neither 
its pressure nor its superheat is known. However, its specific enthalpy is 
1096.6 Btu/lb, the value of h g at 80°F, according to Table 1 of the Keenan 
and Keyes tables. 

10:7. Definitions. In psychrometry, 1 a number of special terms are 
used to aid in the description of the state of the atmospheric-air mixture. 
Among those which we shall use are specific humidity, relative humidity, 
dew point, and dry-bulb and wet-bulb temperature. Although these terms 
are framed for application to the atmospheric-air mixture, they may also 
be used to good advantage in the description of the condition of any gas- 
vapor mixture. 

Specific humidity co is the ratio of the mass of the water vapor to the 
mass of dry air in a given volume of the mixture. It may thus be 
expressed mathematically as 

M s 

OJ = , -p 

M a 


in which the subscript s refers to the water vapor (steam) and a to the dry- 
air component of the mixture. These occupy a common volume, so that 
M s v s = M a v a , or Ms/M a = Va/Vs. Treating both as perfect gases, 
v a = RaTa/Pa = R a T m /Pa, and v s = RsTs/Ps = RsT m /Ps. Substituting 
in the equation that defines co, 


Ms v a Ra Ps 53.3 Ps Q 

" = M a _ Js ~ Us K ~ SKS P~ a ~ ' 


( 10 : 20 ) 


in which p, and p a are the partial pressures of the water vapor and the air,' 2 
respectively. 

1 The science that deals with the behavior of mixtures of air and water vapor. 

2 The term air is here, and in following pages, used to refer to the mixture of gases, 
not including water vapor, which was given that name in Chap. 9. It is sometimes 
called dry air, for emphasis. Equation (10:20) may be applied to gas-vapor mixtures 



216 


BASIC ENGINEERING THERMODYNAMICS 


The maximum specific humidity possible for atmospheric air at a given 
temperature and total (mixture) pressure is that associated with the 
saturated mixture. For saturated-atmospheric-air mixtures, p s is the 
saturation pressure of steam that is equivalent to mixture temperature. 

The relative humidity <f> is the ratio of the density of the water vapor in 
the mixture to its density in a saturated-atmospheric-air mixture at the 
same temperature, or 


= — V JL — RsTm/Pg = Ps = Ps 
* ~ l/v g Vs RsTJPs Pg p g 


( 10 : 21 ) 


in which p g is the saturation pressure of the vapor that corresponds to the 
temperature T m of the mixture [see Eq. (10:19)]. If the mixture is 
saturated, p s has its highest possible value for a given mixture tempera¬ 
ture and is equal to p g . The relative humidity of a saturated mixture is 
therefore 1. 

When an atmospheric-air mixture, the water-vapor component of which 
is superheated steam, is cooled at constant pressure, its volume and tem¬ 
perature decrease and the superheat is gradually removed from the water 
vapor. During this process, the composition of the mixture does not 
change (its specific humidity is constant), and the partial pressures of both 
air and water vapor are (like that of the mixture as a whole) constant. 
Finally, a temperature is reached at which all of the superheat has been 
removed from the water-vapor component, and it is a saturated vapor. 
The mixture is also saturated, and any further cooling will result in 
progressive condensation of the vapor component. The temperature at 
which this condensation begins is called the dew point of the original 
mixture. 

The statement that the partial pressure of the vapor is the same for the 
vapor as the saturation pressure corresponding to its dew point has been 
made above. This may be shown to be the case, based on the perfect-gas 
relation. The ratio M s /M m is constant during the cooling process that 
brings the mixture to its dew point. Then, for the mixture at any point 
in that cooling process, 

P s __ MsRsT 8 /V 8 _ MsRsTJVm _ M s R s 
Pm M m R m TJV m M m R m T m /V m M m R m 


other than atmospheric air if “gas” replaces “air” and “vapor” replaces “water 
vapor.” An equivalent change in the subscripts used in the equation from s to v 
and from a to “gas” might be appropriate, in that case. The constant 0.622 would 
change, of course, to the ratio R ea , B /R v . In giving Eq. (10:20) a more general applica¬ 
tion, in this manner, care should be observed that the pressures of the mixture com¬ 
ponents are low enough so that the perfect-gas relations, on which the equation is 
based, reflect the behavior of these components with sufficient accuracy. 









MIXTURES OF GASES AND VAPORS 


217 


P m is constant, and it follows, since R s /R m must be constant as long as the 
proportions of the mixture components do not change, that P s is constant 
during the process; thus the partial pressure of the water vapor in a mixture 
is its saturation pressure corresponding to the dew-point temperature. This 
permits the substitution of this saturation pressure equivalent to the dew¬ 
point temperature for p s in Eq. (10:21), thereby defining relative humidity 
on a new basis. 

If the process discussed above is carried to temperatures below the dew 
point, condensation will have the effect of reducing the water-vapor con¬ 
tent of the atmospheric-air mixture. This is one method by which the 
dehumidification of atmospheric air may be accomplished. The rate at 
which heat must leave the mixture to decrease the temperature a specified 
number of degrees will increase with the passing of the dew-point tem¬ 
perature, since the enthalpy of vaporization must be removed from the 
weight of vapor condensed in addition to the heat that must be removed 
from the remaining mixture to reduce its temperature. Thus the short¬ 
term temperature variation (as from day to night) to be expected in damp 
climates is normally considerably less than if the climate is dry. 

If atmospheric air at a relative humidity less than 1 is brought into 
intimate contact with liquid water, some of the water will evaporate, with 
a resulting increase in the humidity of the mixture. Heat must be sup¬ 
plied to effect the vaporization of the liquid and comes from the mixing 
components, reducing their temperature. Advantage may be taken of 
this effect either to reduce the temperature of the atmospheric-air mixture, 
as in air conditioning by humidification, or to lower the temperature of 
the water, as when a cooling tower is used to provide a supply of cooled 
water. The process is called adiabatic saturation since, if the water and 
the atmospheric-air mixture are considered to constitute a single system, 
there has been no heat flow and because the limit to which the process may 
proceed is a saturation state for the mixture. 

The process of adiabatic saturation is used for the experimental investi¬ 
gation of the humidity of the atmospheric-air mixture. Two thermom¬ 
eters, one of which is specially prepared by covering its bulb with a wetted 
wick, are placed side by side in a suitable frame and exposed to a stream 
of air. The dry-bulb thermometer will record the temperature of the air; 
its reading is called the dry-bulb temperature. Evaporation from the wick 
of the wet-bulb thermometer produces a saturated-air mixture in immediate 
contact with its bulb and a correspondingly lower temperature called the 
wet-bulb temperature. The amount by which the wet-bulb temperature is 
lower than the dry-bulb temperature is called the wet-bulb depression and 
is a qualitative measure of the capacity of the mixture for containing 
more water vapor, and thus of its dryness. Reference to standard tables 
or charts gives the corresponding relative humidity. 


218 


BASIC ENGINEERING THERMODYNAMICS 


Example 10-7A. Find, for atmospheric air having a total pressure of 14.7 psia, a 
dry-bulb temperature of 80°F, and a relative humidity of 60 per cent (0 = 0.6), 

(a) the partial pressure of the water vapor, (b) the dew point, (c) the superheat of the 
water vapor, ( d ) the specific humidity, and (e) the enthalpy of the water vapor per 
pound of dry air in the mixture. 

Solution: 

(a) 0 = 0.6 = p s /Po, where p g is 0.5069 psia, the saturation pressure at 80°F; p s = 
(0.6)(0.5069) = 0.304 psia; p a = 14.7 - 0.304 = 14.396 psia. 

(b ) p s = saturation pressure at the dew-point temperature. Comparing in Table 1 
of the steam tables, the dew point is found to be 64.8°F. 

(c) The water vapor is superheated 80 — 64.8 = 15.2°F. 

( d ) 03 = 0.622 {p s /pa) = 0.622(0.304/14.396) = 0.0131 lb water vapor per pound of 
dry air. 

(e) The specific enthalpy of the superheated water vapor is the same as that of 
saturated water vapor at the same temperature of 80°F, or 1096.6 Btu/lb (see Art. 
10:6). Therefore H s = M s h s = (0.0131)(1096.6) = 14.4 Btu per pound of dry air 
in the mixture. 

Example 10-7 B. In the combustion of a fuel gas, to which reference is made in 
Example 10:3, the weight of water vapor formed per pound of the dry products of 
combustion is 0.118 lb. (a) When these products are cooled to 100°F at a pressure 
of 14.7 psia, what weight of water vapor has condensed? ( b ) The weight of the dry 
products is 19 lb per pound of fuel gas burned. How does the weight of water con¬ 
densed compare with the weight of fuel burned? 

Solution: 


(a) The products can be assumed to be saturated. At 100°F, p s = p g = 0.9492 psia. 
Then p ga8 = 14.7 — 0.9492 = 13.75 psia (note that p ga8 is used since the dry portion 
of the mixture is not air). 


JE. — El- = 0.602 ( = 0.0415 lb per pound dry products 

Rs Pgas 85.8 p gas \ 13.75 / ^ J * 


Weight condensed = 0.118 — 0.0415 = 0.0765 lb 
weight of water condensed (19) (0.0765) 


( b ) Ratio 


weight of fuel burned 


1 


water per pound dry products 
= 1.45 lb per pound of fuel 


If the difference between the dry products of combustion and dry air had been ignored, 
the corresponding answers would have been (a) 0.0751 lb; (6) 1.43 lb. 


10:8. The Psychrometric Chart. The solution of engineering problems 
that are concerned with atmospheric air is facilitated by the use of a chart 
which plots the specific humidity as ordinates vs. the dry-bulb tempera¬ 
ture as abscissas. This is called a psychrometric chart; it is shown in 
skeleton form as Fig. 10:3, and a detailed psychrometric chart is included 
in the Appendix. 

A psychrometric chart may be constructed for any total mixture pres¬ 
sure by applying the equations of the preceding articles; Fig. 10:3 and the 
chart in the Appendix are based on a total mixture pressure of 14.7 psia, 
standard atmospheric pressure. The first step in construction is the 
location of the saturation (</> = 1.0) curve, the locus of maximum specific 








MIXTURES OF GASES AND VAPORS 


219 


humidity corresponding to each temperature at the given total pressure. 
This may be accomplished by applying Eq. (10:20) substituting the 
saturation pressure corresponding to the dry-bulb temperature as p s and 
with p a — p m — p s . The </> = 1.0 line develops when the results of these 
computations are plotted and the points connected. 



The lines of constant relative humidity are next located; this may be 
done by applying Eq. (10:21), remembering that the saturation pressure 
of the water vapor at the dew-point temperature may be substituted for 
p s in that equation. In Fig. 10:3, point A represents the assumed condi¬ 
tion of an atmospheric-air mixture, having a dew point at D. Then 
4>a = Psa/Poa — Pqd/Pqb- Conversely, it is possible to locate point A so 
that <£ will have a certain value by finding the point D on the saturation 
curve at which the saturation pressure of the vapor is the equivalent frac¬ 
tion of the saturation pressure at B and placing A vertically below B and 
horizontally to the right of D. 

Line AC represents a process of adiabatic saturation and is therefore a 
line of constant wet-bulb temperature. For saturated air, there is no 
wet-bulb depression, and the dry- and wet-bulb temperatures at point C 
are the same. Since the process is adiabatic, with no external work, Eq. 
(10:17) may be applied. Thus, if point A is to be located on a line of con¬ 
stant wet-bulb temperature passing through C, 























220 


BASIC ENGINEERING THERMODYNAMICS 


( h aA + WAh gA ) T (c cc cOA)h/ A = (hac T wchgc) (10:22) 

Enthalpy of mixture Enthalpy of water Enthalpy of mixture 
at A to saturate at C 


It has here been assumed that the water to saturate was supplied at the 
temperature of the mixture at A. Note also that, though the water- 
vapor portion of the mixture at A is really superheated steam, its specific 
enthalpy is denoted as h gA , the enthalpy of the saturated vapor at tem¬ 
perature t A ‘, the justification for this substitution was explained in Art. 
10:6. The second term of the equation constitutes what would be called, 
in mathematics, an infinitesimal, since the difference co c — u A is a very 
small fraction of the total weight of the mixture and hf is small as com¬ 
pared with h g . This term is usually ignored and the equation written in 
the form 

haA + c OAh 0 A hac T wchgc (10:23) 

Thus, with the benefit of the simplification made possible by this approxi¬ 
mation, lines of constant wet-bulb temperature become lines of constant 
mixture enthalpy. Accordingly, the lines of constant wet-bulb tempera¬ 
ture may be extended to register on a scale of mixture enthalpies, as shown 
in Fig. 10:3. In supplying values for h g and h a , the standard practice is 
to use steam-table values for h g and to assume a datum temperature of 
0°F for the air portion of the mixture. Thus the enthalpy of the mixture 
at A is 


H mA — h aA + WAh gA — (1)(0.24) (t A — 0) -f- G>Ahg A — 0.24Ci + a 0Ah gA 

(10:24) 


The psychrometric chart in the Appendix shows the specific humidity 
in terms of grains (1 lb = 7000 grains) of water vapor per pound of dry 
air. The volume of an amount of the mixture that contains 1 lb of dry 
air is also shown. This may be determined if it is remembered that the 
volume of the mixture is the same as that of the dry air, and we may write 


V 

Y rnA 


V aA 


53.3 T a 
Va A 


53.3T a 
Pm ~ Ps A 


(10:25) 


Example 10:8. The steps in the construction of a psychrometric chart similar to 
that in the Appendix are as follows: Construct (a) the saturation, or </> = 1.0, line, 
( b) the constant relative-humidity curves, (c) the lines of constant wet-bulb tempera¬ 
ture, (d) the enthalpy scale, and (e) the lines of constant volume. To illustrate the 
methods used in carrying out these steps, assume the total mixture pressure to be 
14.7 psia, and calculate (a) the specific humidity of saturated air at 70°F, ( b ) the 
specific humidity of air at 80°F having a relative humidity of 60 per cent, (c) the 
specific humidity of air at 90°F dry-bulb and 70°F wet-bulb temperature, ( d ) the 
enthalpy of the mixture having a wet-bulb temperature of 70°F, and (e) the volume 
of the mixture, per pound of dry air, at 80°F, 60 per cent relative humidity. Compare 
the results with data read from the psychrometric chart in the Appendix. Note 




MIXTURES OF GASES AND VAPORS 


221 


that this comparison may be made only because the total mixture pressure is also 
14.7 psia for that chart. 

Solution: 


(a) This calculation was made in Example 10:6. The specific humidity is 0.0157 lb 
water vapor per pound of dry air. The value as read from the chart in the Appendix 
is 110 grains, or ruw o = 0.0157 lb. 

(b) This calculation was made in Example 10:7A, where the specific humidity was 
shown to be 0.0131 lb. This agrees with the chart reading of 92 grains. 

(c) Lines of constant wet-bulb temperature connect states having the same enthalpy 
of the mixture. For the saturated mixture, the wet- and dry-bulb temperatures are 
identical. The enthalpy of the mixture at the given state is therefore the same as for 
saturated air at a dry-bulb temperature of 70°F. This was shown to be 34 Btu in 
Example 10:6. We may write, for the assigned state, 

H m — 34 = ha + C ah, = (0.24) (90) + *>(1100.9) or w = 0.0113 lb 


Note that h g at 90°F has been substituted for the specific enthalpy of the (superheated) 
steam in the mixture. The chart reading is 78.5 grains, which checks closely. 

(< d ) The enthalpy of the mixture at a wet-bulb temperature of 70°F is the same as 
that for a saturated mixture at a dry-bulb temperature of 70°F. This has been calcu¬ 
lated as 34 Btu in Example 10:6. This agrees with the value shown on the chart. 

( e ) The volume of the mixture is the same as that of its components, or 


RaT m _ (53.3) (540) 

Pa (14.396) (144) 


13.9 ft 3 per pound of dry air in mixture 


The value of p a was calculated in Example 10:7A. The chart reading is about 
13.89 ft 3 . 


10:9. Engineering processes for atmospheric air are typically steady- 
flow in character, and Eq. (10:16) may be applied in the solution of the 
engineering problem. The basic processes may be classified as heating or 
cooling at constant pressure (ADE of Fig. 10:3), humidification or dehumidi¬ 
fication of saturated air ( DE), adiabatic saturation (humidification) (AC), 
and adsorption, or adiabatic dehumidification (AF). 

Heating at Constant Pressure. This process could be demonstrated by 
passing the air over closed coils containing a warm fluid. The path on the 
psychrometric chart is horizontal (DA). 

Cooling at Constant Pressure. This process has been discussed in the 
preceding article. The same method can be used as in heating, except 
that the coils are filled with a cold fluid. The path followed is AD and, 
if the process continues to temperatures below that at the dew point, DE. 
This part of the path also represents a dehumidification of saturated air. 

Humidification of Saturated Air. Path ED could be traced by washing 
the air with warm water. 

Adiabatic saturation has also been discussed above. It results when the 
air is washed with water at its own temperature. Path AC is followed. 

In practice, the engineering process may be a combination of two or 
three of these basic processes. For example, if a large excess of chilled 




222 


BASIC ENGINEERING THERMODYNAMICS 


water is used to wash the air, the path traced will be a compromise 
between A CD and AD down to the dew point and will thereafter proceed 
downward along the saturation curve to E. Thus dehumidification of 
the mixture may result from washing. Or a portion of the air may be 
cooled along path ADE and then remixed with the untreated portion, 
which has been by-passed around the cooling coils. By means such as 
this, the condition of the air may be changed to correspond to any desired 
point on the psychrometric chart. 

The purpose of the engineering process may be to adjust the condition 
of the air, either for comfort or industrial use or, as in the case of the cool¬ 
ing tower, to cool the water with which the air is washed. Warm water 
enters at the top of the cooling tower and is sprayed downward into a 
rising stream of air. Atmospheric air enters at the foot of the tower and 
rises, owing either to convective effects or to the positive action of fans, 
against the downward flow maintained by the particles of liquid water 
because of their greater density. A small part of the water evaporates, 
increasing the humidity of the air and being carried off with it at the top 
of the tower. The balance of the water is cooled, largely by this evapora¬ 
tion, and may be drawn off at the bottom of the tower. The final tem¬ 
perature of the water may be less than the dry-bulb temperature of the 
entering air, depending on the relative rates of flow of air and water 
vapor, but cannot be less than its wet-bulb temperature. 

Example 10:9A Four thousand cubic feet per minute of air at 80°F dry-bulb,70°F 
wet-bulb temperature is washed in a spray washer which is supplied with chilled 
water at 42°F. The air leaves the washer saturated at 50°F, and this is also the 
final temperature of the washing water, (a) What weight of cooling water must be 
supplied per hour? ( b) What is the refrigeration (cooling) load in Btu per hour? 
(c) What volume of air leaves the washer per minute? ( d ) What weight of water is 
removed from the air per hour? 

Solution: 

(a) From the chart, at 80°F dry-bulb and 70°F wet-bulb, H m = 34 Btu, V m = 13.89 
ft 3 , co = 94 grains. For saturated air at 50°F, H m — 20.3 Btu, V m = 13.0 ft 3 , 
co = 54 grains. All values are per pound of dry air in the mixture. The specific 
enthalpy of the cooling water at entrance to the washer is 10.05 Btu and, at exit, 
18.07 Btu/lb (hf at 42 and 50°F, respectively). The weight of dry air handled is 

4000 occ u / • 

Yg = 288 lb /min 

The washing process is an adiabatic no-work steady-flow mixing process. The 
kinetic energy and elevation terms of Eq. (10:16) may be disregarded and Eq. (10:17) 
employed. Substitution in this equation will be made with M< t representing the 
weight of the cooling water supplied per pound of dry air in the entering atmospheric 
mixture. Since the outgoing weight of cooling water must be greater than its initial 
weight by the weight of water vapor condensed from the air, 



MIXTURES OF GASES AND VAPORS 


223 


Q4 _ ZA 

Moi = Mix -1 - 7000 — = M* 1 0 0057 

M^ + H mi = Mojiox + H mo or 1^(10.05) + 34.0 = (M h + 0.0057) (18.07) + 

20.3 

Solving, 

Mix = 1.695 lb of cooling water supplied per pound of dry air in mixture. This 
corresponds to a flow of (1.695) (288) (60) = 29,300 lb/hr. 

(6) The refrigeration load, based on the heat that must be removed from the water to 
recool it to 42°F, is 29,300(18.07 - 10.05) = 235,000 Btu/hr. This could also have 
been calculated as the enthalpy loss of the air, or (288) (60) (34.0 — 20.3) = 237,000 
Btu/hr. The discrepancy between the two answers results from neglecting, in the 
second computation, the enthalpy of the water that was condensed from the air. 

( c ) Volume of air leaving washer = (288) (13.0) = 3750 cfm 

(d) Water removed from air = (0.0057) (288) (60) = 98.5 lb/hr 

The difference in the two answers to part b was approximately 2000 Btu/hr. This 
difference is, more accurately, (98.5)(18.07) = 1780 Btu/hr. 

Example 10:9B. If, in Example 10. 9A, the air issuing from the washer had been 
remixed with 2000 cfm of air that had been by-passed around the washer, determine 
the properties of the resulting mixture. 

Solution: 


Weight of dry air by-passed = - = 144 lb/min 

lo.oy 

■tt, i f + + (144) (34.0) + (288) (20.3) 0 . 0 _ , 

Enthalpy of resultant mixture = - 144 -f 288 —' = P er P 0unci 


of dry air 

Its specific humidity 


(144) (94) + (288) (54) 
144 + 288 


67.3 grains per pound of dry air 


The calculated enthalpy is taken to the chart and found to correspond to a wet-bulb 
temperature of about 57.5°F. The intersection of this wet-bulb-temperature line 
with a specific humidity of 67.3 grains gives a dry-bulb temperature of about 60°F. 
Since the enthalpy and specific-humidity scales are linear, this intersection will lie 
on a line joining the states of the uniting mixtures on the psychrometric chart. Other 
properties of the resultant mixture may be read from the chart as V m = 13.3 ft 3 ; 
</> = 0.87; dew point = 56°F. 

Example 10:9(7. Water enters the top of a cooling tower at 110°F and is cooled to 
70°F as it falls to the bottom. Air enters the bottom at 75°F and a relative humidity 
of 40 per cent and leaves at the top at 105°F, relative humidity 95 per cent. Calculate 

(a) the weight of water cooled per pound of dry air passing through the tower and 

( b ) the per cent of cooling water lost by evaporation. 

Solution: 


(a) At 105°F, (f) = 0.95, the partial pressure of the water vapor is (0.95) (1.1016) = 
1.048 psia. The specific humidity of the air as it leaves is 

coo = 0.622 (jjy^fo 4 s) = 0,0477 lb per P ound of dr y air 

H mo = (0.24) (105) + (0.0477) (1107.3) = 78.0 Btu per pound of dry air. 

At entrance, the specific humidity, as read from the chart, is 51.5 grains, or 0.0074 lb 
per pound of dry air, and the enthalpy is 26.3 Btu . 1 

1 The General Electric Company psychrometric chart shown in the Appendix calls 
this the total heat instead of the enthalpy. This conforms to an earlier practice, now 







224 


BASIC ENGINEERING THERMODYNAMICS 


The specific enthalpy of the water is 77.94 Btu/lb at the top and 38.04 Btu/lb at 
the bottom of the tower. 

Equation (10:17) will apply and, for unit weight of dry-air flow, 

Mq(77.94) + (1)(26.3) = [M ix - (0.0477 - 0.0074)](38.04) + (1)(78.0) 


from which 


= 1.26 lb of water cooled per pound of dry air through the tower 
(b) Per cent of water lost by evaporation = 26^ ^^ ) (100) = 3.2 per cent 


Problems 

1. A mixture has the following proportions by volume: CO 2 —13 per cent; O 2 — 
6 per cent; N 2 —81 per cent. Its pressure is 25 psia and its temperature 100°F. The 
internal energy and the entropy are arbitrarily zero for each gas existing separately at 
14.7 psia and 32°F. Find (a) the apparent molecular weight and the gas constant of 
the mixture, (6) the gravimetric analysis, (c) the specific internal energy, enthalpy, 
and entropy of the mixture, ( d) the partial pressure of each constituent, (e) the specific 
volume of the mixture, and (/) the ratio k for the mixture. 

2. A mixture at 25 psia and 100°F has the following gravimetric analysis: C0 2 — 
50 per cent; 0 2 —40 per cent; helium—10 per cent. Find (a) its apparent molecular 
weight and gas constant, ( b ) its specific volume, (c) the mole fractions of its com¬ 
ponents, ( d ) their partial pressures, and ( e ) the exponent k for reversible adiabatic 
expansion. 

3. Substitute the following mixtures (percentages by volume) for the mixture of 
Prob. 1, and find the same quantities: (a) N 2 50 per cent and O 2 50 per cent; (b) C0 2 
50 per cent and 0 2 50 per cent; (c) He 50 per cent and 0 2 50 per cent. 

4. Substitute the following mixtures (percentages by weight) for the mixture of 
Prob. 2, and find the same quantities: (a) N 2 50 per cent and 0 2 50 per cent; (6) C0 2 
50 per cent and 0 2 50 per cent; (c) He 50 per cent and 0 2 50 per cent. 

5. A mixture consists of 0.2 mole of nitrogen, and the balance is helium. Its total 
pressure is 100 psia, and its volume is 20 ft 3 . The partial pressure of the helium is 
40 psia. What is the temperature of the mixture? 

6. The combustion equation for methane is CH 4 + 20 2 —> C0 2 + 2H 2 0. What 
is the specific volume of a mixture of methane with just enough oxygen to support 
complete combustion if the pressure of the mixture is 14.7 psia and its temperature 
is 70°F? If c v for methane is 0.45 Btu/(lb)(°F), what is the exponent A; for reversible 
adiabatic compression for this mixture? 

7. In Prob. 6, change the mixture to a mixture of methane with just enough air to 
furnish the oxygen for its complete combustion. If air is a mixture of 3.29 lb of 
nitrogen per pound of oxygen, calculate the gas constant of the mixture. What is the 
ratio of the specific heats, &? 

8. Tank 1 has an internal volume of 4 ft 3 , tank 2 a volume of 6 ft 3 . The pressure 
and temperature of the contents of each tank are 14.7 psia and 32°F. The two tanks 
are connected, and their contents are allowed to form a mixture at a common tempera¬ 
ture and pressure. Describe the composition of the resulting mixture in terms of the 
mole fractions of its components, state the pressure and temperature of the mixture 


almost never seen. The confusion that might be caused in the student’s mind by the 
use of “total heat” is obvious. The term enthalpy was devised for the purpose of 
eliminating that confusion. 




MIXTURES OF GASES AND VAPORS 


225 


that results from adiabatic mixing, and calculate the change of entropy that takes 
place, if the tanks originally held, respectively, (a) nitrogen and oxygen, ( b ) CO 2 and 
O 2 , (c) helium and O 2 ; ( d ) if both tanks contained oxygen. In each case, is the mixing 
process reversible? 

9. The same as Prob. 8, except that the pressure differs in the two tanks before 
they are connected. For tank 1 it is 20 psia; for 2 it is 15 psia. The temperature 
remains at 32°F initially in both tanks. For gas combinations as described in Prob. 8, 
answer the same questions. 

10. Same as Prob. 8, except that the temperature differs in the two tanks before 
they are connected. For tank 1 it is 50°F; for tank 2 it is 32°F. The pressure is 
initially 14.7 psia in both tanks. For gas combinations as described in Prob. 8, 
answer the same questions. 

11. Work Example 10:4, changing the amounts of each gas from 1 lb. to 1 ft 3 in 
each compartment. Answer the same questions. 

12. As they pass through the mixing valve on the way to the cylinder of an internal- 
combustion engine, a steady stream of methane (see Probs. 6 and 7) at 70°F is mixed 
with a steady stream of air at 90°F. The proportionate rates of flow are such that 
the air will contain just enough oxygen for complete combustion of the methane. 
What is the resultant temperature of the mixture? Assume the mixing to be adia¬ 
batic and the velocities of the methane and the air entering and of the mixture leaving 
the valve to differ negligibly. 

13. In Prob. 12, let the velocity of the air entering and the mixture leaving the 
valve be 20 fps, while the methane enters at 400 fps. Heat enters through the walls 
that confine the flow between the two sections at the rate of 50 Btu per pound of 
methane entering the mixture. What is the temperature of the mixture? 

14. A gas mixture is composed of 20 per cent C0 2 , 30 per cent 0 2 , and 50 per cent 
N 2 by weight. In steady flow through a heat exchanger, its pressure decreases from 
20 to 15 psia and its temperature from 220 to 130°F. Differences in velocity and in 
elevation between entrance and exit are negligible. Find the amount of heat trans¬ 
ferred per pound of the mixture. What is the change of entropy per pound? 

15. Ten thousand pounds per hour of water at 80°F enters an open feedwater 
heater, where it comes in contact with steam at 14.7 psia and a quality of 0.90. If 
the resulting mixture is withdrawn at 210°F at a rate such that the liquid level in the 
heater is constant, how many pounds leave the heater per hour? 

16. A mixture is composed of equal percentages by weight of carbon dioxide and 
helium. It is compressed reversibly and adiabatically from 14.7 psia, 70°F, to a final 
pressure of 75 psia. How much work is required per pound of the mixture? Per 
pound of the mixture, what are the individual entropy changes of each constituent? 

17. The same as Prob. 16, except that the percentages are equal by volume. 

18. Work Example 10:6 for the following temperatures and total pressures of the 
saturated-atmospheric-air mixture: (a) 90°F and 14.7 psia; ( b ) 90°F and 10 psia; 
(c) 70°F and 10 psia. What conclusions do you draw as to the effect of temperature 
on the mole fraction of water vapor in the saturated-air mixture? As to the effect of 
total mixture pressure? 

19. In Prob. 7 assume the combustion of the methane to be complete, i.e., to be 
represented by the combustion equation of Prob. 6. What are the weights of, respec¬ 
tively, C0 2 , N 2 , and H 2 0 in the products per mole of methane burned? When these 
products have a total pressure of 14.7 psia and a temperature of 70°F, what fraction 
of the total weight of H 2 0 which was formed must have condensed? What are the 
partial pressures of the gaseous components of the products at the stated temperature 
and total pressure? 


226 


BASIC ENGINEERING THERMODYNAMICS 


20. Based upon the steam-table convention as to the reference level from which 
enthalpy is measured, find the specific enthalpy of steam at the states described as 
follows: (a) p =0.10 psia, t = 60°F; (6) p =0.10 psia, t = 80°F; (c) p = 0.20 psia, 
superheated 30°F. 

21. Water is allowed to evaporate into an “atmosphere” of pure hydrogen until 
a saturated mixture results. If the temperature and pressure of this mixture are 
70°F and 14.7 psia, what is its specific humidity? What are the mole fractions of its 
components? If the “atmosphere” is carbon dioxide, what are the similar values? 

22. In Example 10:7 A, change the total pressure to 10 psia, and find the same 
properties. 

23. Find the specific volume of steam at the states listed in Prob. 20. 

24. Atmospheric air has a total pressure of 14.7 psia, a temperature of 50°F, and a 
relative humidity of 40 per cent. Find (a) the partial pressure of the water vapor; 
( b ) the dew point; (c) the superheat of the water vapor; ( d ) the specific humidity; 
(e) the enthalpy of the water vapor per pound of dry air in the mixture. 

25. The total pressure of the air in a heated room is 14.7 psia. At the center of the 
room the temperature of the air is maintained at 70°F and its relative humidity at 
50 per cent, but near an exposed wall the temperature is lower owing to a flow of 
heat outward through the wall. How low may be the temperature at the wall without 
resulting in a condensation of moisture on that surface? 

26. In Example 10:7A, calculate the wet-bulb temperature. Change the total 
pressure to 10 psia (as in Prob. 22), and again calculate the wet-bulb temperature. 
Compare your calculations for the two pressures. Do you obtain different answers 
for the two pressures, and if so, what causes the difference? 

27. A mixture of hydrogen and water vapor has a pressure of 14.7 psia and a tem¬ 
perature of 70°F. Its relative humidity is 60 per cent, (a) What is the partial 
pressure of the water vapor? ( b ) What is the dew point? (c) What is the specific 
humidity? ( d ) What is the wet-bulb temperature? Repeat for a mixture of carbon 
dioxide and water vapor at the same pressure, temperature, and relative humidity. 

28. Atmospheric air at 14.7 psia and 80°F has a wet-bulb temperature of 50°F. 
Calculate (a) its specific humidity; (b) its dew point; (c) its relative humidity. Change 
the total pressure to 10 psia, and calculate the same properties. 

29. Using the psychrometric chart, solve Probs. 24, 25, 26, and 28. Check the 
solution of Example 10:7A. Can the chart be used for the solution of Prob. 22? For 
the solution of Prob. 27? 

30. A saturated mixture of hydrogen and ammonia vapor has a pressure of 180 psia 
and a temperature of 0°F. What are the partial pressures of its components? What 
are their percentages by weight in the mixture? 

31. Given air at a dry-bulb temperature of 80°F and having a dew point of 70°F, 
read the following from the psychrometric chart: (a) relative humidity; ( b ) wet-bulb 
temperature; (c) specific humidity; (d) enthalpy; (e) volume. What restriction, if 
any, applies to the use of these values? 

32. Air at 88°F and 40 per cent relative humidity is humidified adiabatically as it 
passes through a washer. What is its final temperature as it leaves the washer as 
saturated air? Use the psychrometric chart. What is the change in specific 
humidity? 

33. Air at 88°F and 40 per cent relative humidity is passed through cooling coils 
in which its temperature is lowered until the air becomes saturated. What is the final 
temperature? How much heat is removed per pound of dry air? Use the chart. 

34. Fourteen hundred cubic feet per minute of air at 85°F dry-bulb temperature 
and 50 per cent relative humidity is mixed with 200 cfm of air at 54°F dry-bulb tem¬ 
perature and 40 per cent relative humidity. Based on values read from the chart, 


MIXTURES OF GASES AND VAPORS 227 

find the resulting dry-bulb temperature, wet-bulb temperature, and relative 
humidity. 

35. IIow many cubic feet per minute of air at 70°F dry-bulb temperature and 
50 per cent relative humidity must enter a cooling tower to cool 100 gal of water per 
minute from 85 to 70°F, assuming that the air is saturated at 80°F as it leaves the 
top of the tower? 

36. An evaporative cooler is used to cool a residence in the summer. The cooler 
draws 3400 cfm of outside air with a dry-bulb temperature of 92°F and a relative 
humidity of 20 per cent. While in the cooler, the air is sprayed with water until, 
when discharged into the house, its relative humidity has been increased (adiabati- 
cally) to 90 per cent. How many pounds of water are evaporated into the air per 
minute? What is the final dry-bulb temperature of the air? 

37. Ten thousand cubic feet per minute of air at 60°F dry-bulb temperature and 
40 per cent relative humidity is passed over a cooling coil. Assuming that the air 
leaves the coil at 55°F dry-bulb temperature, how many Btu are removed per hour 
by the coil? 

38. Find the weight of dry air that must be drawn hourly through a cooling tower 
to cool 60,000 lb of water per hour from 105 to 75°F if the air enters at 70°F and with a 
relative humidity of 30 per cent and leaves at 100°F and a relative humidity of 95 per 
cent. What percentage of the water is evaporated? 

39. The absolute humidity of moist air is the weight of water vapor per cubic foot 
of air. Construct a psychrometric chart on which the absolute humidity is plotted 
as the ordinate vs. dry-bulb temperatures as abscissas, plotting the following loci: 
100 per cent relative humidity, 50 per cent relative humidity. Note that these loci 
are independent of the total pressure of the mixture. Assuming the total pressure 
to be 14.7 psia, plot the line which represents a constant wet-bulb temperature of 
70°F. 

40. Construct a skeleton psychrometric chart similar to Fig. 10:3 but based on a 
total atmospheric pressure of 24.89 in. Hg (equivalent to an altitude of 5000 ft above 
sea level). Take the following steps in order: (a) Calculate data for saturation 
(100 per cent relative humidity) curve at 10°F intervals, and plot. (6) Based on the 
dew-point temperature, calculate data for the 50 per cent relative humidity line, and 
plot, (c) Calculate necessary data, and plot a line of 70°F wet-bulb temperature. 
(d) Determine the locus of conditions for which the total volume of the standard unit 
mixture is 16 ft 3 , and plot this locus on the chart. 

Symbols 


c p specific heat at constant pressure 
c v specific heat at constant volume 
C a constant 
g acceleration of gravity 
h enthalpy of unit mass 
H enthalpy of a system 
J proportionality factor 
k ratio of the specific heats, c p /c v 
m molecular weight; weight of 1 mole 
M mass; mass rate of flow 
n a constant 
p pressure, psi 

P pressure, psf; pressure in general 
Q rate of heat flow 


228 


BASIC ENGINEERING THERMODYNAMICS 


R gas constant 
s entropy of unit mass 
S entropy of a system 
T absolute temperature 
u internal energy of unit mass 
U internal energy of a system 
v specific volume 

V volume of a system 

V velocity 

W rate of work delivery 

x mole fraction of a component in a mixture 
z elevation 

Greek Letters 

<j> relative humidity 
co specific humidity 

Subscripts 
a air 

e natural logarithmic base 
g of the saturated vapor 
i incoming 
rn mixture 
o outgoing 
p constant pressure 
s steam; water vapor 
v constant volume 


CHAPTER 11 




STEADY FLOW OF FLUIDS—THE TURBINE 


11:1. Introduction. A study of the behavior of fluids in steady flow is 
useful in the analysis of optimum performance of turbines and other 
devices that utilize a fluid in steady flow, in the design of these units, in 
the design of pipes and ducts for conveying fluids, and in the measurement 
of fluid flow. All of these are purposes that may be of maj or importance 
to the engineer. Only those aspects which are concerned with the turbine 
and similar devices will be treated in the present chapter. 

All real fluids have the property called viscosity in greater or in less 
degree. The flow of a viscous fluid can never be reversible, but it will be 
our policy to discuss first the limiting, or reversible, process, ignoring 
viscosity. The discrepancies between ideal and actual performance will 
then be noted. For example, there will be no drop in pressure along a 
horizontal pipe or duct of constant cross-sectional area which carries a 
nonviscous fluid in adiabatic flow, and the fluid will issue from the passage 
in exactly the same state as when it enters; nor will any work be required 
for its transport. This varies widely from the conditions encountered in 
real flow through such lines, and these conditions cannot be predicted 
until viscosity is taken into account. 

The basic tools for the analysis of the behavior of the fluid in steady flow 
are the steady-flow energy equation [Eq. (3:5)] and the continuity equa¬ 
tion of steady flow [Eq. (3:8)]. For ease of reference, these equations will 
be repeated here. 


hl + ihg + 1 + lQz “ K + 


iV , 22 , iWt 

2Jg + J + J 


M i — M 2 — 


AiVi A2F5 


or 


M = 


AV 


Vi 


V2 


V 


(3:5) 

(3:8) 


These two equations apply to the steady flow of fluids whether that flow is 
reversible or irreversible. They may also be used, with certain limita¬ 
tions, when the flow is not strictly steady in character (see Art. 3:9). 

The turbine is not a heat engine; as the prime mover, it may form a 
part of the steady-flow heat engine (see Art. 3:8 and Fig. 3:4). In the 
turbine, a dual operation is carried out. The fluid is first given increased 
velocity by reducing its pressure as it passes through a constricted pas¬ 
sageway, called a nozzle , and this kinetic energy is then removed, by 

229 







230 


BASIC ENGINEERING THERMODYNAMICS 


means of moving blades, or buckets , to reappear as shaft work; the velocity 
may be partly generated in the passages between the buckets themselves. 
The turbine principle may be applied whether the fluid is a vapor, a gas, 
or a liquid. The steam turbine is perhaps most familiar to the average 
engineer, and its operating principle is sketched in Fig. 11:1. The figure 
shows a very simple type of the steam turbine, which is built in a multi¬ 
tude of forms, but serves to illustrate the general principle of operation of 
all turbines, whether steam, water, or gas. The analysis of optimum 
turbine performance is carried out by first analyzing the flow through the 



Fig. 11:1. DeLaval steam turbine. 


nozzle and then the flow through the bucket which converts the kinetic 
energy of the jet into shaft work, these flows being treated as reversible 
(frictionless) in character. 

11:2. The Reversible Adiabatic Nozzle. The flow through the nozzle 
is assumed to be adiabatic because the brief time required for passage 
from entrance to exit gives little opportunity for appreciable heat flow. 
Because limiting, or reversible, flow is the subject of our study at present, 
the flow is also isentropic. This establishes a relationship between h\ and 
hz in Eq. (3:5) since it requires that the entropies Si and s 2 , at entrance and 
exit, respectively, shall be equal. The differences in elevation between 
entrance and exit are insignificant, and the elevation terms in the equa¬ 
tion may be dropped. No external work is performed in the nozzle, and 
we may rewrite Eq. (3:5), as it applies to the reversible nozzle, as 

V 2 2 Vi 2 

2Jg ~ Wg = ( hl ~ h2) ‘ 

in which the enclosure of the enthalpy difference and the use of the sub- 























STEADY FLOW OF FLUIDS—THE TURBINE 


231 


script s reminds us of the isentropic character of the flow. In the nozzle, 
the object is to increase the velocity of flow, and V 2 > F\; a decrease of 
enthalpy must take place, and h\ > h 2 . When the velocity ratio F 2 /F 1 
is large, it is often possible to drop V\ 2 /2Jg from this equation as an 
approximation; for example, if V 2 /Vi = 20, then V 2 2 = 400Fi 2 and the 
error invited amounts to only one-fourth of 1 per cent. If this approxi¬ 
mation is employed, Eq. (11:1) simplifies to 


F 2 = VJgih - = 223.8[(/q - hi) J* 


( 11 : 2 ) 


Assuming that the state of the fluid at entrance to the nozzle is known, 
including the initial pressure pi, and that we have adequate information 
as to the relation between the properties of the fluid (as, perhaps, in the 
form of an equation of state or a table of properties), it is possible to 


establish the state at exit if p 2 is known, since one other property ( s 2 ) 


corresponding to that state is also known. These calculable properties 
will, of course, include h 2 and v 2 . We are now in a position to. apply Eq. 
(11:1) or Eq. (11:2) to the calculation of V 2 . Next, from Eq. (3:8), 



and the cross-sectional exit area required for a given rate of fluid flow can 


be computed by substituting the previously determined values of v 2 and 


V 2 in this equation. 

The flow is isentropic at all sections of the nozzle, and if it becomes 
desirable to calculate the cross-sectional area of the nozzle at some inter¬ 
mediate station a , where the pressure p a has some known value between 
Pi and p 2 , the same methods may be applied to the calculation of A a . 

11:3. The Liquid Nozzle. It may be shown that, during the isen¬ 
tropic process, 



This is the basis of a convenient method of calculating the change of 
enthalpy of an incompressible fluid during isentropic flow. Liquids are 
compressible in such small degree that we may safely consider v to be 


constant in connection with the liquid nozzle. Thus, substituting in Eq. 


( 11 : 1 ), 

V 2 2 F 1 2 


= (h-i — h 2 ) s = — 


I, 


2 


dh = - 



2 


dP = j (Pi - P t ) 


2Jg 2 Jg 


or 



(11:4) 











232 


BASIC ENGINEERING THERMODYNAMICS 


indicating that the increase of kinetic energy of the liquid, as it flows 
through the nozzle, is proportional to the drop in pressure. 

In Eq. (11:3), M and v are both constants for liquid flow and a con¬ 
tinuous decrease in nozzle cross-sectional area in the direction of flow is 
evident if the velocity is continuously to increase. This is a converging 
nozzle such as is illustrated in Fig. 11:2. The shape of the walls here 
shown is such as to make the rate of velocity change per unit length ol 
channel larger near the entrance, where the velocities are low, and smaller 
near the exit. This is found to give good results in the real nozzle, for 
which frictional effects must be considered. The reversible adiabatic 



Fig. 11 : 2 . Converging nozzle. Fig. 11 : 3 . The venturi. 

nozzle may be designed to give equal increments of velocity, of kinetic 
energy, or equal pressure drop per unit of nozzle length, as desired. The 
real nozzle should be shaped to give a minimum of irreversibility (friction) 
in flow of the fluid; its proportions are based on the results of experimental 
observation. 

If the liquid, having reached the exit section of the converging nozzle 
shown in Fig. 11:2, is then caused to enter a diverging passageway which 
ultimately reaches, in its cross section, the inlet area of the converging 
nozzle, as shown in Fig. 11:3, a venturi is formed. The section of this 
venturi downstream from the throat of the venturi , its smallest section, is 
called a diffuser. The pressure rises in the direction of flow in a diffuser, 
and if the flow is adiabatic and reversible and the areas at entrance to and 
exit from the venturi are equal, the pressure of the liquid at exit from the 
diffuser will have been built up to equal that at entrance to the nozzle and 
all other properties at these two sections will also be the same. The 
viscosity of a real fluid will produce frictional effects that will make 
reversible flow impossible, and the pressure at exit from the diffuser will 
fall somewhat short of equaling that at entrance to the nozzle section of 
the venturi. The comparison of the pressure loss between these two 













STEADY FLOW OF FLUIDS—THE TURBINE 


233 


sections, when equivalent flow conditions are maintained, furnishes one 
means of comparing the viscosities of fluids. 

Example 11:3. A water nozzle is to be designed to discharge 10,000 lb of water per 
minute at a velocity of 150 fps and to connect to a line having an internal diameter 
of G in. Assuming discharge is to the atmosphere and neglecting friction, calculate 
(a) the diameter ot the nozzle at its exit and (6) the pressure and velocity of the water 
in the line as it approaches the nozzle. 

Solution: 


Mv 2 _ 10,000 

V 2 ~ (60) (62.4) (150) 


0.0178 ft 2 = 2.56 in. 2 


Diameter at exit = D 2 



(h) V = — 1 = (10,000) (144) 

A, (60) (62.4) (0.7854) (36) 


13.6 fps 


1.81 in. 


Substituting in Eq. (11:4), 

150 2 13.6 2 144 , 

60 ~ 60 = 60 (?>1 ~ 14 7) ° r Vl = 170 psia 


11:4. The Perfect-gas Nozzle. For the perfect gas, 


(hi - h 2 ) s = c p (Ti - T 2 ) s = J (T 1 - T 2 ) s 

k RI\\ 
k - 1 J [ l 

Making this substitution in Eq. (11:1), we have 



(11:5) 


TV _ TV 
2 g 2 g 


Jc p (Ti 




RTi 



( 11 : 6 ) 


As the pressure of a gas decreases during flow through a nozzle, the specific 
volume will increase, at first slowly and then more and more rapidly. 
This means that the increase of velocity in a gas nozzle is relatively much 
more rapid than in the flow of an incompressible liquid and suggests the 
feasibility of applying Eq. (11:2), as follows: 





RI\ 


1 - 



223.8[c p (T 1 - T 2 ) 8 ]i (11:7) 


In the flow of a gas through a nozzle, the velocity and the specific vol¬ 
ume increase as the pressure decreases in the direction of flow. In apply¬ 
ing Eq. (3:8) to determine the shape of the nozzle, let us consider two 
successive sections across the passage, a and 6, as in Fig. 11:4; the direc¬ 
tion of flow is as indicated, and b is the downstream section. Let the 
specific volume and velocity of the fluid at section a be denoted as v and V, 
























234 


BASIC ENGINEERING THERMODYNAMICS 


respectively, and the higher values of these properties at section b will be 
designated as v + dv and V + dV. Then, from Eq. (3:8), 

Ab _ Vb V a _ v + dv V _ 1 + dv/v 1 

T a ~~ 


V, 


( 11 : 8 ) 


V b v V + dV 1 1 + dV/V 

It is evident from Eq. (11:8) that if dV/V is larger than dv/v then Ab/A a 
< 1 and the downstream section will be smaller. This is the condition 
a £ observed in the first section of a nozzle 

when the fluid enters at low velocity and 
indicates that this section of the nozzle 
should converge. As the pressure drops 
lower and lower, the rate of increase of 
specific volume becomes greater while the 
corresponding rate of increase of velocity 
becomes smaller; in later stages of the 
expansion, dv/v may therefore become 
larger than dV/V, and Eq. (11:8) indicates 
that nozzle areas must increase in the di¬ 
rection of flow. When dv/v = dV/V, the 
nozzle will have reached its smallest area, or throat. Note that this will 
be the throat of a nozzle rather than a venturi, since the pressure will con¬ 
tinue to decrease in the diverging section, although the passage may 
resemble a venturi in appearance. 

The state of the fluid reached at the throat of a reversible adiabatic 
perfect-gas nozzle may be investigated by applying the equality dv/v = 
dV/V. For the isentropic process, we may write 



Fig. 11:4. Adjacent nozzle sec 
tions. 


Pv k — p iVi k or v — pp/k Vl p-i/k an( j — 


dv _ / — 1 
v \ k 


-1 

k 


pp^pC-k-lWk dp 


pp/kvJH.-k-V/k ) (Pj-l/A^-ipi/*) dP = 


-dP 

kP 


(11:9) 


Applying Eq. (11:7) to a section of the channel where the pressure is P, 


V 2 = 2g k 

= 2g k- 1 
dV 2 VdV 


k - 1 
k 


RTi 


P 1 V 1 — 2 g 



C k-D/k 


= 2g 


k 


k - 1 

Pl 1/ ^iP(fc-i )/ A: 


PlVl 


1 - 


(?) 


(k — l)/k 


k - 1 
2^Pi 1/ ^ 1 P- 1/fc dP 


V 


2 V 2 




k 


k - 1 


PlVl 


(?) 


(*-!)/*' 


dP 


k 


k - 1 


[Pfk-D/k _ p(fc-i)/A:] 


( 11 : 10 ) 






























STEADY FLOW OF FLUIDS—THE TURBINE 235 


At the throat, dv/v — dV/V. Designating the pressure at the throat 
as P 0 , 


dP 

kP 0 


■Pr 1/k dP 


2 /Y fc - 1)/l - 2 A -P 0 ik ~ l),k 

k — 1 k — 1 


[Eqs. (11:9) and (11:10)] 


from which 


or 


and 


fcp 0 (*-iv* = 2 _A_ p^k-D/k _ 2 k «/* 

(fc + 2 ^ = 2 P/*-')'* 

/p Xtt-D/c _ 2/c/(fc _ ^ 2 

VV fc(fc + l)/(fc - 1) k + 1 



( 11 : 11 ) 


If the value of ft for the monatomic gas (1.67) is substituted in Eq. 
(11:11), the ratio of throat to initial pressure will be found to be about 
0.49. For air and the diatomic gases, the ratio is 0.53. For more com¬ 
plicated molecules, higher ratios will result. This ratio is called the 
critical-pressure ratio for gas flow. 

In the flow of a gas through a converging nozzle such as that of Fig. 
11:2, the pressure at the smallest section of the channel will be equal to 
the pressure on the exhaust side of the nozzle (the final pressure, pi) if that 
pressure is higher than the critical throat pressure, as calculated by apply¬ 
ing Eq. (11:11). In that case, this section of smallest area would be the 
last section that would play any part in the expansion of the fluid and 
would therefore be the exit section. Accordingly, as illustrated in Fig. 
11:5a, conditions at that point are designated by the subscript 2. If the 
nozzle is circular in cross section, the jet will issue from the nozzle in 
cylindrical form, as shown. If, as in sketch b of the figure, the exit pres¬ 
sure p 2 is equal to the critical throat pressure, the flow will again issue in 
the form of a cylinder from the smallest section of the nozzle. This sec¬ 
tion may now, however, be properly called the throat, and the subscript 0 
can be used to designate conditions at that point. The mass rate of flow 
will be somewhat greater than for case a because of the larger drop in pres¬ 
sure and temperature. Lastly, as the exhaust pressure becomes less than 
the critical throat pressure, the pressure at the throat will remain constant 
at p 0 , the pressure drop from p 0 to p 2 taking place after the nozzle exit has 
been passed. In this part of the expansion, dv/v > dV/V, and the jet 














236 


BASIC ENGINEERING THERMODYNAMICS 


expands in the exhaust space, as shown in Fig. 11 :5c. The mass rate of 
flow will be determined by the flow through the smallest section and 
therefore by the drop in pressure and temperature between the entrance 
and this section. As the final pressure p 2 decreases, the pressure at the 



Fig. 11:5. How through a converging nozzle for various values of the ratio P 2 /P 1 . 


0 


smallest section will decrease with it until p 2 equals p 0 , the critical throat 
pressure. Any further decrease of p 2 will not change the pressure at the 
throat, and consequently the mass rate of flow through the nozzle will 

reach a maximum as p 2 becomes equal to p 0 
and will thereafter remain constant at this 
maximum for any further decrease of the ex¬ 
haust pressure. When p 2 < p 0 , the nozzle is 
called an expanding nozzle. To reduce fric¬ 
tional effects in real flow, expanding nozzles 
should be designed with walls that continue 
beyond the throat section, diverging until the 
exit area is sufficient to meet the requirements 
of the flowing jet as it reaches the final pres¬ 
sure p 2 . 

In Fig. 11:6 is shown a converging-diverg¬ 
ing nozzle. Also, this nozzle is acting as an 
expanding nozzle, as is indicated by the in¬ 
crease in v and V beyond the throat in the 
chart at the lower part of the figure. These 
increases could have resulted only from a 
decreasing pressure in the diverging section 
of the nozzle, and therefore the channel is not a venturi, for the venturi 
always includes a diffuser, with rising pressure in the direction of fluid 
flow. Yet the same channel may become a venturi under certain con- 



Fig. 11:6 
nozzle. 


The expanding 






















































STEADY FLOW OF FLUIDS—THE TURBINE 


237 


ditions, as when p 2 is equal to or greater than the critical throat pres¬ 
sure in gas flow or when the flowing fluid is an incompressible liquid. 

The chart of Fig. 11:6 plots the specific volume and velocity of the gas 
against the distance traveled along the nozzle axis. Examination of the 
equation that connects the pressure and specific volume during an isen- 
tropic expansion will indicate that the specific volume will increase, at 
first slowly and then more and more rapidly, with respect to a unit 
decrease in pressure. The velocity, starting from practically zero, will 
increase at first very rapidly and later at a decreasing rate with respect to 
a unit pressure drop, as the fluid passes through the nozzle; this may be 
confirmed by returning to Eq. (11:7). If the scales to which v and V are 
measured on the graph are adjusted to values such that the same distance 
measures both v 0 and Vo, the two curves will be tangent at the throat of 
the nozzle, as shown in Fig. 11:6. 

The velocity at the throat of a gas nozzle may be calculated from Eqs. 
(11:7) and Eq. (11:11), as follows: 


Fo 2 = 2 g 


k 


k - 1 


RT 1 


-m 


(fc—l)/fc 




[Eq. (9:40)] 


To 
Ti " 

Also, 


k -1 

Po\ k 

.Pi, 


k k — 1 


,k — 1 k 


fc + 1 


k T 1 


R To — P o^o 


and, substituting in the equation for V 0 ' 2 above, 


tv = 2 » fcTr i p ° v ° k -Ar ! 1 - 


2 


k + 1 


= 2 gJ^P 0 v 0 k+lk ~ 1 


k - 1 


2 k + 1 

= kgP Q Vo 


or 


Vo = (kgP 0 Vo) h 


( 11 : 12 ) 


Referring to texts in the field of physics, it is found that this expression 
is given as the velocity of sound in a gas at the pressure P 0 and having the 
specific volume Vo- Thus the velocity attained at the throat in isenti opic 
flow through an expanding nozzle is the velocity of sound in that gas 
under the conditions which exist at that point; this is the maximum 
velocity which can be reached in the converging section ol a nozzle, the 
velocity downstream from the throat exceeds V o and is greater than the 
velocity of sound for the expanding isentropic nozzle. 

If a gas, already traveling at a velocity in excess of the velocity of sound 
in that gas, enters a converging passage, its pressure will rise and its 

















238 


BASIC ENGINEERING THERMODYNAMICS 


velocity decrease in isentropic flow and the passage will act as a diffuser. 
From the point where, because of this diffuser effect, the velocity drops to 
the velocity of sound, the passage must diverge if the action is to continue 
to be that of a diffuser and to further build up the pressure and reduce the 
velocity. Thus the converging-diverging nozzle of Fig. 11:6 may become 
a diffuser over its entire length, and the implications of reversibility are 
emphasized. Conversely, if the flowing gas is to expand to lower pres¬ 
sures and is already traveling at the velocity of sound or above, a diverg¬ 
ing passage must be provided. 



(a) Nozzle effect 


\\\\\\\ 

))>))))) 

/////// 

(b) No nozzle effect 
Fig. 11:7. Flow through turbine blading. 


Nozzles need not, of course, be circular in cross section. The effect of 
a converging nozzle is often secured, in turbine design, by causing the 
fluid to flow through restricted passages between blades, or vanes , as 
shown in Fig. 11:7a. Here the exit area is less than that at entrance, and 
the fluid (traveling at a velocity less than that of sound) will leave it at 
increased velocity. When the purpose is merely to change the direction 
of flow without giving the effect of a nozzle, blades shaped as in Fig. 11 :7b 
are used. 


Example 11:4. What areas at the throat and exit of an air nozzle are required for 
the expansion of 3600 lb /hr from a pressure of 100 psia, temperature of 300°F, to a 
final pressure of 14.7 psia? Assume frictionless flow. 

Solution: 


Po / 

Pi \ 

= r. g) 


2 y/(*-D 

k+1) 


( 2 \ 14 / 0.4 

= ( 2^. ) = 0.528 or p 0 = 52.8 psia 




= (760)(0.528)°- 4/1 - 4 = 633°R 



STEADY FLOW OF FLUIDS—THE TURBINE 


239 


t 2 = Ti 


^p_ 2 y k ~ i)/k 



388°R 


Vo = 223.8[c p (7 1 1 - ToW = 223.8(0.24(760 - 633)]* = 1235 fps 
V 2 = 223.8(0.24(760 - 388)]* = 2110 fps 


v 0 = 


v 2 = 


A o = 


RT o _ (53.3) (633) 

Po (52.8) (144) ~ 
RT 2 _ (53.3)(388) 

P 2 (14.7)(144) “ 
Mvo _ (3600) (4.44) 
V 0 (60) (60) (1235) 


4.44 ft 3 /lb 
9.78 ft 3 /lb 
= 0.0036 ft 2 


0.518 in. 2 at throat 


A 2 = 


Mv 2 

V 2 


(3600) (9.78) 
(60) (60) (2110) 


= 0.00463 ft 2 = 0.668 in. 2 at exit. 


11:5. The Vapor Nozzle. The principles that govern the reversible 
flow of vapors are the same as for the flow of a perfect gas, with the excep¬ 
tion that the perfect-gas relation cannot be employed and we must turn 
to tables and charts connecting the properties of the particular vapor in 
order to design a suitable nozzle for its expansion. For that purpose, 
Eqs. (11:1) and (11:2) will be found useful if the flow is isentropic, and 
Eq. (11:3) will also apply. Expansion to a certain critical-pressure ratio 
will, as for gas flow, be accompanied by the attainment of velocities equal 
to the velocity of sound in the vapor and by maximum flow rates for a 
given initial state. To expand to pressures less than this critical throat 
pressure requires a diverging nozzle passage, so that a throat is as charac¬ 
teristic of the expanding vapor nozzle as it is for the expanding perfect-gas 
nozzle. 

The mass rate of flow through vapor nozzles is evidently again depend¬ 
ent upon the area of the smallest section. For maximum flow, dv/v 0 = 
dV/Vo at this point, as before. Equation (11:11) cannot, in strict accu¬ 
racy, be applied to find the critical-pressure ratio Ro/pi, since the deriva¬ 
tion of that equation was based on the perfect-gas relation. But the 
isentropic process may be followed for the vapor, and, by trial and error, 
the critical pressure for a given expansion may be determined as the pres¬ 
sure reached when dv/v = dV/V. The ratio of this pressure to the 
assumed initial pressure gives the critical-pressure ratio for this expansion. 
A large number of such calculations are, of course, necessary before any 
conclusions may be drawn as to a general value of the ratio. However, 
experience has shown that Eq. (11 ill) will hold very closely for the vapor, 
although k for the vapor is no longer the ratio of the specific heats, as for a 
gas. To avoid confusion, we shall designate it as n. In determining the 
value of n for use in this equation as it applies to the vapor, the pressures 
and volumes at two states having the same entropy may be determined 
and the relation P iVi n = P 2 v 2 n written, as for the perfect gas; from this 
relation, n may be computed. For superheated steam, the value of n, 
determined in this manner, quite uniformly approximates 1.31, and, for 











240 


BASIC ENGINEERING THERMODYNAMICS 


saturated steam, the corresponding value is 1.13. When these values 
are substituted in Eq. (11:11), the corresponding critical-pressure ratios 

are 

For steam initially superheated, 

2? = 0.55 
Pi 

For steam initially saturated, 

^ = 0.58 
Pi 

The Moilier chart will be found useful in the design of the vapor nozzle. 
The state of the vapor at entry to the nozzle is located as point 1 of Fig. 

11:8. The expansion in the re¬ 
versible adiabatic nozzle is isen- 
tropic, and a vertical line (ds = 0) 
through point 1 cuts the exhaust 
pressure at point 2, locating the 
state of the vapor at exit. From 
the data thus obtained, the veloc¬ 
ity of the vapor at exit from the 
nozzle and the required area of the 
nozzle at that point for a desired 
mass rate of flow may be calcu¬ 
lated. For an expanding nozzle, 
such as is shown in the figure, 
the required throat area must also 
be determined. The condition at 
the throat is located as the inter¬ 
section of the vertical line previ¬ 
ously drawn with the constant- 
pressure line representing the critical throat pressure. If the vapor is 
steam and the state at entrance is in the superheat region, as illustrated 
in Fig. 11:8, this critical throat pressure p 0 equals 0.55pi; if the steam 
is initially saturated, p 0 is 0.58pi. Other ratios would, of course, 
apply to other vapors and could be determined by applying the 
methods discussed in the earlier part of this article. When the point 0 has 
been located in the manner described above, the properties of the fluid at 
the throat are available and F 0 and A 0 , the throat velocity and throat 
area for the required rate of flow, may be successively calculated. 

The dashed line of Fig. 11:8 represents the effects of friction in the real 
nozzle and will be discussed later. It will be noted that, in accordance 
with the principle brought out in Chap. 6, the nozzle condition line l-0'-2' 



Fig. 11:8. Design of expanding vapor 
nozzle (data of Examples 11:5 and 11:6). 





STEADY FLOW OF FLUIDS—THE TURBINE 


241 


moves toward increasing entropies in reflecting the effects of irreversi¬ 
bility. The shape of the real nozzle expresses the designer’s ideas and 
information as to the form that will give minimum friction in flow; 
simplicity and economy of the manufacturing process must also be con¬ 
sidered. Between entrance and #throat, a gradual convergence is the 
custom, resulting in a well-rounded entry section like that of the con¬ 
verging nozzle of Fig. 11:2. Between throat and exit, the walls are often 
conical in shape. When the nozzle is not an expanding nozzle (when the 
ratio P 2 /P 1 is greater than the critical-pressure ratio), the required effect 
may be obtained by the use of vanes shaped as in Fig. 11:7a. 


Example 11:5. Find the required throat and exit areas for a nozzle to expand 
3600 lb of steam per hour from 100 psia, 360°F, to a final pressure of 2 psia. Assume 
frictionless flow. 

Solution. The steam is initially superheated, and therefore p 0 = (0.55) (100) 
= 55 psia. The Mollier chart is used to obtain the data in the first three columns 
of the table below; in obtaining the tabulated data, the initial state is projected 
vertically (at constant entropy) to cut the lines of throat and exit pressures, and the 
data are read at the points of intersection. If the intersection is in the superheat 
region, the temperature is recorded in the third column; if below the saturation line, 
the quality is set down. 


Section 

V 

h 

t or x 

V 

V 

A 

Entrance 

100 

1206 

360°F 




Throat 

55 

1156 

0.979 

1580 

7.61 

0.695 

Exit 

2 

944 

0.832 

3610 

144.5 

5.76 


The values tabulated in the last three columns of the table are calculated, as below: 


Vo = 223.8(Ai - ho) 1 * = 223.8(1206 - 1156)* = 1580 fps 


F 2 

Vo 

V 2 


Ao 


= 223.8(/n - h 2 )* = 223.8(1206 - 944)* = 3610 fps 
= x 0 v ao = (0.979)(7.787) = 7.62 ft 3 /lb 
= X 2 v 02 = (0.832)(173.73) = 144.5 ft 71b 


Mv 0 

~W> 


(3600) (7.62) 
(60) (60) (1580) 


= 0.00482 ft 2 = 0.695 in. 2 


Mv 2 (3600) (144.5) 
V 2 (60) (60) (3610) 


0.040 ft 2 = 5.76 in. 2 


In this example, the steam was wet steam at both throat and exit, and the condition 
was tabulated in terms of the quality. If it had been superheated at one or both of 
these points, the temperature would have been tabulated and the specific volume 
obtained from Table 3 of the steam tables, by interpolation if necessary. 


11:6. The Real Nozzle. The effects of irreversibility in flow through 
the real nozzle, whether it be designed to expand a liquid, a gas, or a vapor, 
are expressed in terms of the coefficient oj velocity , coefficient of discharge , 



































242 


BASIC ENGINEERING THERMODYNAMICS 


and the nozzle efficiency. The nozzle condition line bends to increasing 
entropies at lower pressures, in accordance with the principles outlined in 
Chap. 6, and follows a curve like l-0'-2' of Fig. 11:8. 

Nozzle efficiency is defined as the ratio of the kinetic energy of the stream 
as it leaves the real nozzle to that developable at exit from a reversible 
adiabatic nozzle which accepts the fluid at the same initial state (including 
the initial, or approach, velocity as one of the properties defining that 
state) and exhausts to the same final pressure. This ideal kinetic energy 
may be computed from Eq. (11:1) and is 

V-2 V 2 

-Ti- = -7T- + J Afc* n . (H:13) 

2 g 2 g 


where Vi, the ideal velocity at exit, corresponds to V 2 for isentropic ex¬ 
pansion, Va is the approach velocity [corresponding to Vi in Eq. (11:1)], 
and the isentropic drop in specific enthalpy in the reversible nozzle 
[(hi — h 2 ) s in Eq. (11:1)] is designated as A h sn . The efficiency may then 
be expressed as 


tv A; 

V a 2 /2g + J Ah 


(11:14) 


where rj n is the nozzle efficiency, a decimal fraction, and V 2 > is the velocity 
at exit from the real nozzle. When the approach velocity is negligible, 1 
its effect may be ignored and Eq. (11:14) simplified to the form 


tV/2</ 

J Ahgn 


(11:15) 


Equation (3:5) applies to the irreversible as well as to the reversible flow 
process and, neglecting approach velocity in flow through the real nozzle, 
that equation shows that tV/2 g = J (hi — h 2 >) and 


V n = 


hi — h 2 ' 
A h S n 


hi — h 2 r 
hi — h 2 


(11:16) 


where the subscripts refer to the corresponding points in Fig. 11:8. The 
efficiency of a well-designed expanding nozzle may be expected to be in 
the neighborhood of 0.90. Frictional effects increase rapidly with the 
velocity of the fluid, and the efficiency of the nonexpanding nozzle may be 
somewhat higher. 

The coefficient of velocity of a nozzle is defined as the ratio of the velocity 
of the fluid as it leaves the real nozzle to its velocity as it leaves the ideal 


1 Calculation will show that it requires an approach velocity of about 224 fps to 
have the effect of adding 1 Btu to the isentropic enthalpy drop Ah sn in the denominator 
of Eq. (11:14). 








STEADY FLOW OF FLUIDS—THE TURBINE 


243 


nozzle which operates under equivalent conditions. This ideal velocity 
is, from Eq. (11:1), 

Vi = (7a 2 + 2Jg A h' n )* 

and 



7y 

(7 a 2 + 2 Jg A h an )* 


(11:17) 


where C Vn is the nozzle coefficient of velocity, a decimal fraction. Neglect¬ 
ing the approach velocity, this becomes 



V y 

(2 Jg A h an )t 


(11:18) 


A comparison of Eq. (11:17) with Eq. (11:15) or of Eq. (11:18) with Eq. 
(11:16) will show, in either case, that 

C„ = Vn * (11:19) 


When the real nozzle is to be designed as an expanding nozzle, the 
throat area must be calculated to give the desired rate of flow. The 
velocities in the converging section of the nozzle are less than those after 
the throat is passed. This causes the effects of friction to be less in 
evidence at the throat than over the entire expansion, and the coefficient 
of velocity at the throat is higher than the coefficient of velocity of the 
nozzle. In fact, it is often assumed that all friction takes place between 
throat and exit of the expanding nozzle, and the entry section is designed 
as a reversible adiabatic nozzle. Based upon the value of the nozzle 
efficiency suggested above, it is evident that the coefficient of velocity to 
be expected in the well-designed expanding nozzle is about 0.95; it cannot, 
of course, exceed 1, since any value in excess of 1 would be a violation of 
Second Law principles. A reasonable value of the coefficient of velocity 
at the throat is about 0.99. Any value less than 1 at this location will 
mean that the velocity of sound will not be attained as the throat is passed 
in the real expanding nozzle and that this velocity will not be reached 
until at least some distance after the diverging section of the channel has 
been entered. 

The coefficient of discharge compares the mass rate of fluid flow through 
the real nozzle with that through the same nozzle when the flow is isen- 
tropic and the same entry conditions and exhaust pressure apply. If the 
nozzle is nonexpanding, the theoretical rate will be based on the exit area 
of the nozzle and is Mi = A 2 V s /v s or, neglecting approach velocity, 

M _ MV S _ MVsn 
Ld ~ Mi ~ J^fi ~ 223.8A 2 (A/i sn )* 


( 11 : 20 ) 






244 


BASIC ENGINEERING THERMODYNAMICS 


where Cd 
M 


v Sn = 


discharge coefficient 

mass rate of flow through the real nozzle 
specific volume of the fluid at the exhaust pressure and an 
entropy equal to that at entrance to the nozzle 
ideal velocity, or velocity attained as the result of isentropic 
flow 

If the nozzle is an expanding nozzle, the theoretical mass rate of flow 
will be based on the throat area A 0 and 


V s = 


C d 


Mv 8nt 

223.8A 0 (A/O* 


( 11 : 21 ) 


where v Snt is the specific volume at the critical throat pressure and an 
entropy equal to that at entrance to the nozzle and A h Snt is the isentropic 
drop of enthalpy from entrance to throat of the reversible adiabatic 
nozzle. 

Since the effect of friction is to slow down the fluid and to increase its 
specific volume above that which would result from frictionless flow (the 
entropy is higher than would result from isentropic flow, and this, at the 
same pressure, means greater volume), it would seem that the discharge 
coefficient, like the coefficient of velocity, could not exceed 1. This is not 
a requirement of the Second Law, however, and we shall shortly investi¬ 
gate flow situations that are accompanied by mass rates of flow higher 
than the theoretical. 


Example 11:6. Assume conditions similar to those of Example 11:5 except that 
the flow is not reversible. The coefficient of velocity at the throat is assumed to be 
0.99 and the nozzle efficiency (entrance to exit) as 0.90. Based upon the same rate of 
flow (3600 lb/hr), calculate the required areas at throat and exit, the discharge coeffi¬ 
cient, and the velocity and velocity coefficient at the exit. Neglect approach velocity 
as in Example 11:5. 

Solution. The velocity at the throat is (0.99) (1580) = 1564 fps. This velocity 
is equivalent to an enthalpy drop from entrance to throat of 1564 2 /(64.4)(778) = 
49 Btu, and the enthalpy at the throat is therefore 1206 — 49 = 1157 Btu. When this 
enthalpy and the pressure of 55 psia are taken to the Mollier diagram, the quality at 
the throat is observed to be 0.98; the specific volume is (0.98) (7.787) = 7.63 ft 3 /lb. 
Then A 0 = (3600)(7.63)/(60)(60)(1564) = 0.00487 ft 2 = 0.702 in. 2 This is the area 
at the throat required to obtain the desired rate of flow. The coefficient of discharge 
for this expanding nozzle is based on Eq. (11:21), 

= Mv st = (3600) (7.62) 

Ao(Ah Snt )l (60) (60) (0.00487) (50)1 

The coefficient of discharge might also have been calculated as the ratio of the throat 
area required to obtain the desired rate of flow if frictionless to that area necessary 
under the assigned conditions, or 0.00482/0.00487 = 0.99. It will be noted that, 
unless the effects of friction are unusually large, the discharge coefficient will be 
approximately equal to the velocity coefficient at the throat (or at the exit, if the 
nozzle is a converging nozzle). 






STEADY FLOW OF FLUIDS—THE TURBINE 


245 


The nozzle efficiency is 0.90. From Eq. (11:16), 
0.90 = / 2 O 6 - 944 0r hr = 970 Btu/lb 


This enthalpy and the final pressure of 2 psia make it possible to read the final quality 
from the chart, as x 2 > = 0.856. The specific volume at exit is v 2 > = (0.856) (173.73) 
= 148.5 ft 3 /lb. Also, 


V 2 

A 2 

C vn 


= 223.8(1206 - 970)5 = 3430 f ps 

_ (3600)(148.5) = 00433 ft* = 6 24 in 3 
(60) (60) (3430) U U 66 U b - 24in - 

= V J = 0.905 = 0.949 


Checking, 

C Vn = mi = 0.95 


11:7. Pressure Variation in the Convergent-Divergent Nozzle. Figure 
11:9 is adapted 1 from a report of tests made on a convergent-divergent 
steam nozzle at various ratios of exhaust to supply pressure. It plots the 


P 

P, 



£ 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 

q Relative distance from nozzle entry ^ 

kj 


Fig. 11:9. Pressure variation in a convergent-divergent nozzle. 


ratio of the pressure at successive stations along the length of the nozzle 
to the pressure at which superheated steam was supplied the nozzle, for 
various ratios of exhaust to initial pressure, p 2 /pi. These pressures were 
measured at a series of small holes in a tube which extended along the axis 
of the nozzle. 

1 From Binnie and Woods, Proc. Inst. Mech. Engrs., 138 (1938). 



























246 


BASIC ENGINEERING THERMODYNAMICS 


The nozzle was apparently designed for a pressure ratio of expansion o] 
about 0.2, and curve A of the figure shows the observed pressure variatior 
under that condition. It will be noted that the pressure ratio at the 
throat agrees with the critical-pressure ratio for superheated steam. The 
small rise in pressure on curve A near the 0.7 station is the result of sudden 
condensation that is produced due to a flow condition to be discussed in 
later pages of this chapter; the pressure in the nozzle again decreases after 
this discontinuity has been passed. Curve E records the pressure varia¬ 
tion in the same nozzle when the exhaust pressure is about 87 per cent of 
the supply pressure. For this case, the pressure at the throat, while 
above the critical throat pressure, is less than the exhaust pressure, and 
the diverging section of the channel acts as a diffuser. In calculating the 
discharge coefficient, the theoretical mass rate of flow would be based on 
the assumption that the pressure at the throat was the exhaust pressure, 
and it would therefore be possible for the discharge coefficient to be 
greater than 1. In curve D, where the ratio P 2 /P 1 is about 0.82, the same 
effect is observed, but in this case the throat pressure has declined to its 
critical, or minimum, value and the mass rate of flow through the nozzle is 
probably near the maximum. Although the pressure is still dropping at a 
point slightly beyond the throat, it is doubtful that the velocity of sound 
is attained, because of frictional effects. Curve C, with p 2 /pi equal to 
about 0.7, follows the pressure variation of curve A to a point well 
beyond the throat and then breaks sharply upward from this curve. The 
first sudden rise in pressure (denoted by s on the curve) is caused by a 
standing pressure wave in the nozzle which is called pressure shock. After 
passing this shock wave, the pressure rise becomes more gradual and more 
typical of diffuser action. To enter a shock wave of this kind, the fluid 
must have attained a velocity greater than the velocity of sound. Since 
the pressure is lower at this point than that at exit from the nozzle, the 
later sections of the nozzle must act as a diffuser. But these sections con¬ 
tinue to diverge and can have a diffuser effect only if the velocity of the 
fluid is less than that of sound, as explained in Art. 11:4. The fluid enters 
the shock wave at a velocity higher than that of sound and leaves it at a 
velocity which has suddenly been reduced below that of sound; it is then 
in a condition such that it can utilize the diverging passage as a diffuser. 

A more complete thermodynamic explanation of pressure shock will be 
reserved for a later chapter. It will be shown there that the flow through 
the shock is irreversible and always characterized by decreasing velocities 
and increasing pressures. Being irreversible, it causes a loss in nozzle 
efficiency and is therefore to be avoided in design. The effect on nozzle 
efficiency is most serious when the exhaust pressure is about midway 
between the exhaust pressure of curve D, where the throat pressure first 
reached its critical value, and that of curve A, for which the nozzle was 
designed. 


STEADY FLOW OF FLUIDS—THE TURBINE 


247 


11:8. Metastable Expansion—Supersaturated Steam. The preceding 
article has included an explanation of why the discharge coefficient of the 
converging-diverging nozzle may exceed 1. But even the converging 
nozzle may have discharge coefficients slightly greater than 1, and the 
conditions that make this possible are the subject of our present dis¬ 
cussion. 

In Fig. 11:8 is shown the change in state of the steam as it flows through 
a steam nozzle. If this were a converging nozzle, the expansion could pro¬ 
ceed no further than point 0, but, for the case illustrated, where the steam 
enters the nozzle moderately superheated, the steam would have crossed 
the saturation line in the course of its expansion. If this expansion had 
taken place slowly, the crossing of the saturation line would have marked 
the condition at which the kinetic energy of the molecules is no longer 
sufficient to resist their mutual attraction for one another and small 
groups of molecules would have begun to form nuclei, about which still 
further condensation could proceed. But the expansion, as it proceeds in 
the nozzle, is exceedingly rapid and is accompanied by sweeping tempera¬ 
ture changes. Here and there, molecules will form into small collections 
but, until the group reaches a certain critical size, will tend to again sepa¬ 
rate. This tendency toward reevaporation of the minute droplet is a 
result of the surface tension of the liquid in producing the effect of a taut, 
rubberlike skin over the surface of the droplet. The tension in this skin 
accounts, because of the small radius of curvature of the surface, for a 
disparity between the pressure of the liquid and of the vapor molecules, 
and the former find it easier to escape than do the latter to become a part 
of the group of liquid molecules. When the droplets reach a certain 
critical size, which is smaller for lower steam pressures and correspond¬ 
ingly higher specific volumes, this effect vanishes and condensation begins. 
Condensation, when it starts, is rapid, and a dense cloud of liquid particles 
makes its appearance suddenly at a point in the nozzle well beyond that 
at which condensation would have begun if stable equilibrium had been 
maintained. Yellott 1 observed this phenomenon through the glass walls 
of a specially prepared nozzle. The steam, between the point where it 
crosses the saturation line and that at which stable expansion is resumed 
with the sudden formation of the cloud of liquid droplets, is called super¬ 
saturated steam. Its temperature is less than the saturation temperature 
corresponding to its pressure. It is said to be in a state of metastable 
equilibrium, meaning that a disturbance of finite magnitude is required to 
restore it to completely stable equilibrium. It might be compared to the 
equilibrium of a ball resting in the bottom of a shallow depression at the 
top of a cone. For small disturbances, the ball is in stable equilibrium 
and will return to the bottom of its hollow; a disturbance sufficient to 

1 J. I. Yellott and C. K. Holland, “Condensation of Steam in Diverging Nozzles,” 
Trans. ASME, 59 , (April, 1937). 


248 


BASIC ENGINEERING THERMODYNAMICS 


move it over the rim of the depression will bring about a sweeping change. 
The necessary disturbance, in the case of supersaturated steam, is the 
formation of drops of the critical size. This may be the result either of 
the accidental accumulation of a sufficient number of molecules or of a 
continuation of the expansion to pressures low enough so that very small 
collections will suffice. At a pressure of 6 psia, a collection of about 12 
molecules will form a drop of critical size and this group will have a radius 
of curvature of about 1.4 X 1CF 9 ft. 

Yellott and Holland (see footnote, page 247) found that the limit of 
supersaturation in the flow of steam through a converging-diverging noz¬ 
zle (the point at which the cloud of liquid particles was suddenly formed) 
was about 60 Btu below the saturation line on the Mollier diagram. If 
the steam has passed the throat of the nozzle in the form of supersaturated 
steam, conditions, at the throat, including the mass rate of flow, will be 
affected. To investigate these conditions, it is customary to extrapolate 
the equilibrium relations for the superheated steam into the region below 
the saturation line. It will be remembered that n during isentropic 
expansion of superheated steam is about 1.31, for saturated steam around 
1.13. When the two isentropic expansion curves, Pv L31 = C and Pv 1AZ 
= C, are compared on a PV diagram (see Fig. 9:3), it will be observed 
that the expansion as supersaturated steam will give greater densities at 
equal pressures. Neither the pressure nor the velocity at the throat 
will be greatly different for the two types of expansion. This means that 
a larger mass of supersaturated steam will pass the throat than would be 
possible if the steam were wet. Thus, discharge coefficients larger than 1 
are possible if supersaturation is present at the throat of a nozzle. If the 
steam is highly superheated at entrance to the nozzle, the throat may be 
passed before the steam reaches the saturation line and becomes super¬ 
saturated; in this case, supersaturation could not account for flow rates 
above the theoretical. 

The sudden accumulation of the cloud of liquid particles at the limit of 
supersaturation is accompanied by a small rise of pressure of the steam as 
the steam adjusts to a state of stable equilibrium. It is this effect which 
accounts for the discontinuity in curve A of Fig. 11:9, to which attention 
was called in Art. 11:7. 


Example 11:8. In Example 11:5, assume that the steam was supersaturated as it 
passed the throat of the nozzle. Calculate the coefficient of discharge. 

Solution. The flow equations of a gas with n = 1.31 are used to extrapolate the 
expansion of the steam as a dry vapor to throat pressure. The critical-pressure ratio 


Po _ f 2 \»/(»-!) 

pi \n + 1/ 


/ 2 \ 1 . 31 / 0.31 

V2N1J 


= 0.542, or p 0 = 54.2 psia 


Equation (11:7) may be used to calculate the velocity. 



STEADY FLOW OF FLUIDS—THE TURBINE 


249 


m^) r 

= {(64.4) (g|{) (85.8) (820) [1 - (0.542)«w 

= 1605 fps 

/pAV* / l \ 1/1.31 

"-•‘(is) - 4663 (0542) = 7.43ftvib 

M = I*' = (6-60482)(1605) _ „ , b/ Qr 3744 , 

Vo 7.43 ’ 

This corresponds to a discharge coefficient of lipjij = 1.04. 


■«]}* 


11:9. Turbine Staging. Steam may be supplied to the modem central- 
station turbine prime mover at pressures above 1000 psia, and the exhaust 
pressure may be less than 1 psia. When these pressures are taken to a 
Mollier diagram for steam, it is found that the isentropic drop in specific 
enthalpy would approach 600 Btu, and if the expansion were carried out 
in a single bank of nozzles, the theoretical velocity reached by the steam 
would be some 5500 fps. The losses caused by flow friction increase 
rapidly with the velocity after the velocity of sound (about 1500 to 1700 
fps) is passed, as has been brought out in earlier pages of this chapter, 
and even if the reduction of irreversibility due to friction were the only 
gain to be expected, the division of the total drop of pressure and enthalpy 
into steps, by the use of banks of nozzles in series, would have an advan¬ 
tage which could not be ignored. But there are other benefits, several of 
which will develop in the course of the discussion to follow. For example, 
if the expansion is staged, it will be possible to withdraw part or all of the 
steam at pressures intermediate between the pressures of supply and of 
exhaust, should that be desirable for any reason. 

If the total enthalpy drop through the turbine is divided into steps of 
50 Btu or less, the maximum velocity reached by the steam will be less 
than the acoustic velocity and converging nozzles may be used. The 
effect of a converging nozzle may be attained by the use of the blade 
arrangement illustrated in Fig. 11 :7a; nozzles so formed have the advan¬ 
tage of economy in their cost and their space requirements. 

If the series of pressure drops occur only in stationary nozzles, with no 
essential drop in pressure through the moving buckets, the staging is of 
the impulse type; if, on the other hand, these pressure drops take place in 
both stationary and moving blades, the resulting arrangement is called 
reaction staging. A pure impulse or a pure reaction stage is an idealization 
never found in the real turbine; all real turbine stages have some of the 
elements of impulse and some of the characteristics of reaction design. 
For example, blades of the type sketched in Fig. 11:75 are used as the 
buckets in the impulse stage; because of fluid friction, the velocity at exit 
tends to be less than at entrance, but this is impossible, according to the 






250 


BASIC ENGINEERING THERMODYNAMICS 


continuity equation, since the area at exit is no larger than that at 
entrance. The pressure must therefore drop at least slightly across these 
moving buckets, and this means a corresponding increase of specific vol¬ 
ume of the fluid and, according to Eq. (3:8), an increased velocity at exit. 
It is evident that this requirement of increased velocity has the effect of 
still further increasing the pressure drop across the bucket. The staging 
of the real turbine is classified as impulse or reaction according to the 
characteristics which predominate in its design. The same turbine may 
utilize impulse stages over a part of its over-all pressure and enthalpy drop 
and reaction staging for the balance, depending upon the relative advan¬ 
tages of each in high- and low-pressure ranges. It is not appropriate to 
the purposes of this text to go into the details of turbine design, and we 
shall content ourselves with an examination of the basic types of staging 
and their general characteristics from the thermodynamic viewpoint. 

11:10. Impulse Staging. A series of four impulse stages is diagramed 
in Fig. 11:10; the pattern of the pressure, enthalpy, and velocity changes 
of the fluid as it moves through these stages is charted below. It is here 
assumed that the bucket velocity is the same in each stage; this would be 
the case, for example, if the blade-circle diameter (the diameter of a circle 
cutting the mid-height of each blade in a ring of blades) of each row of 
moving buckets was the same and the disks, or rotors, on which these 
buckets are mounted, were keyed to the same shaft. Each stage begins 
at the entrance to a row of stationary nozzles and ends when the fluid is 
ready to enter the next row of nozzles. The enthalpy drop is the same for 
each stage, and, assuming the flow through the buckets to be frictionless, 
this drop is effected entirely in the stationary nozzles. To obtain these 
equal enthalpy drops, the pressure drops through the high-pressure stages 
must be greater than through succeeding stages; there is no pressure drop 
across the buckets for frictionless flow. The absolute velocity (the veloc¬ 
ity relative to the stationary nozzles) of the fluid reaches a maximum at 
exit from each row of nozzles and a minimum at exit from the moving 
buckets of each stage; the maxima and minima are the same for each 
stage under the conditions we have assumed. Because the mass rate of 
flow through each stage is the same, as is the kinetic energy of the fluid at 
entrance to and exit from each row of moving blades, each stage develops 
the same power. 

Some features of the design of the impulse turbine are immediately 
apparent. Since there is only a very small drop in pressure across the 
moving buckets (zero drop for the reversible case illustrated), leakage 
over the top of these blades is negligible and it is unnecessary to provide 
close clearances between them and the turbine casing. At a given instant, 
work is performed on those buckets which are directly in the path of the 
jet as it issues from the stationary nozzles; the rest of the buckets are idle. 


STEADY FLOW OF FLUIDS—THE TURBINE 


251 


In the lower pressure stages, the specific volume of the fluid is greater, and 
larger passages must be provided for flow through both nozzles and mov¬ 
ing buckets. This increased area is provided both by increasing the 
height of both moving and stationary blades and by increasing the propor¬ 
tionate number of moving buckets that are in action at one time; in the 
first few stages the nozzles cover only a portion of the periphery. 



Fig. 11:10. Enthalpy, pressure, and velocity variation through an impulse turbine. 

In Fig. 11:11 is shown a vector diagram of the velocities through one of 
the stages of Fig. 11 :10; the vector diagram for all of its stages will be the 
same under the assumptions made. In the construction of Fig. 11 : 11 , it 
is assumed that the flow through the moving bucket is reversible (friction¬ 
less) so that the velocity V 2 , relative to the bucket at entrance, is equal to 
the velocity V 3 , relative to the same bucket at exit. The bucket entrance 
and exit angles /3i and /? 2 are equal. The absolute velocity of the fluid at 
exit from the stationary nozzle is designated as V 1 , and this vector makes 
an angle a (the nozzle angle) to the plane of rotation of the bucket. The 
absolute velocity of the bucket is indicated as Vb, and the final (absolute) 
exit velocity from the stage is F 4 . The vectorial subtraction of Vb 




















































252 BASIC ENGINEERING THERMODYNAMICS 

from Vi gives F 2 . The entry angle of the bucket is such that the jet may 
enter tangentially to the surface of the bucket on which it is to act, and 
is therefore established from the direction of V 2 . When friction is 
ignored and the velocity relative to the bucket is the same in magnitude 
at exit as at entrance, f3 2 , the exit angle of the bucket is made equal to /3i, 
as in the diagram. The vectorial addition of Vb and V 3 gives the absolute 
exit velocity F 4 . 



Fig. 11:11. Velocity diagram—impulse stage (maximum work). 


It is convenient to resolve the various velocities into their components 
parallel and perpendicular to the plane of bucket rotation. The parallel 
component will be designated by the subscript x, and the perpendicular 
component by y; thus V\ x = V\ cos a and V\ y = Fi sin a. Referring to 
Fig. 11:11 and considering the direction of Vb as positive, we may write 
the following relations: 

V 2x = Vu_— V b = Fi cos a — V b 
V u = 7 V 2x 

V 4 X = Vs x H~ Vb — — V 1 cos a 2Vb 


The force on the bucket may be calculated as equal to the change of 
momentum of the fluid in the direction of bucket movement and the work 
per pound of fluid flow as the product of this force and Vb. Thus 


W per lb = FVb 


(f 2j - FJ Vb = 2(V 1 cos « - V b )Vb 

g g { ' 


For maximum work, 

dW 2 - - 1 

—zr = - (Fi cos a — 2 Vb) =0 or = - Fi cos a (11:23) 

dv b g * 


The maximum work is found by substitution of this value of V b in Eq. 
( 11 : 22 ), or 


W max per lb = - ( 7i cos a 

g 


1 - 


1 - 




^ V 1 cos aj-Vi cos a = cos 2 a (11:24) 














STEADY FLOW OF FLUIDS—THE TURBINE 


253 


It will be noted from Eq. (11:24) that, as the nozzle angle approaches 
zero, total conversion of the kinetic energy of the jet, as it leaves the 
stationary nozzles, into work on the impulse bucket is approached. How¬ 
ever, the component of velocity normal to the plane of blade rotation 
must be large enough to clear the fluid, as it exits from the moving 
buckets, from the path of following buckets. Moreover, except in the 
final stage of expansion, this unremoved kinetic energy is not a thermo¬ 
dynamic loss, for it allows the fluid to enter the next row of nozzles with 
an approach velocity and is thus, in theory, completely recoverable in 
later expansion. Nozzle angles are normally around 20°. 

Equation (11:22) indicates that the work output of the impulse stage 
and, indirectly, the turbine efficiency are quite sensitive to changes in 
rotative speed. For best results, the bucket speed should differ only 
negligibly from that expressed in Eq. (11:23). The impulse turbine is 
designed to operate at constant speed and is not suited to variable-speed 
operation. Equation (11:23) also brings out an additional advantage of 
multistaging. If the entire expansion were effected in a single row of 
nozzles, it has been shown in Art. 11:9 that the velocity V\ might con¬ 
ceivably approach 5500 fps. The optimum bucket speed, which corre¬ 
sponds to this jet velocity and a nozzle angle of 20 deg, is about 2500 fps. 
For a rotor of 4 ft diameter, this corresponds to about 12,000 rpm. This 
would be an impossible speed from the standpoint of design because of the 
excessive centrifugal stresses which would be created in the rotor; more¬ 
over, the utilization of power delivered under this condition of extreme 
rotative speed would be difficult and would invite high losses due to 
mechanical friction and other factors. 

A two-row impulse stage is diagramed in Fig. 11:12, and a vector diagram 
of this stage is presented in Fig. 11:13. Only a single row of stationary 
nozzles is used; the stationary vanes have the function of merely changing 
the direction of fluid flow. In the vector diagram, frictionless flow is 
assumed, and V 2 = V 3 , F 4 = Vs, and V &— IN in magnitude, though not 
in direction. The blade angles are shown as equal at entrance to and exit 
from each row of blades, with the exception of the row of blades that forms 
the stationary nozzles, but these blades become shallower in the direction 
of flow. Approximately half the velocity of the jet is removed in the first 
row of buckets; thus the two-row impulse stage accounts for about three- 
fourths of its power output from the first row of moving buckets, one- 
fourth from the second. For a given bucket speed, jet velocities are 
higher for this two-row stage as compared with the single-row impulse 
stage discussed above, and frictional losses are correspondingly greater, 
but it has the virtue of permitting a larger drop of pressure and enthalpy 
per row of buckets at the expense of somewhat lowered efficiency. For 
the same blade speed, note that V\ for the two-row stage is approximately 


254 


BASIC ENGINEERING THERMODYNAMICS 


double V\ for the single row; this is equivalent to an enthalpy drop four 
times as large. Therefore, based on the same bucket speed, the arrange¬ 
ment of Fig. 11:12 is capable of utilizing about the same enthalpy drop as 
the four impulse stages of Fig. 11:10. Three- and four-row stages are also 




Fig. 11:12. Enthalpy, pres- Fig. 11:13. Velocity diagram—two-row 

sure, and velocity variation impulse stage (maximum work), 

through a two-row impulse 
stage. 

conceivable and would be equivalent, respectively, to 9 and 16 single-row 
stages, but more than two are seldom used because of the mounting fric¬ 
tional losses. Two-row stages are sometimes used for the first drop in 
pressure in a steam turbine, to be succeeded by single-row stages at the 
lower pressures. 

Example 11:10. Assume a reversible single-row impulse stage with an isentropic 
enthalpy drop of 40 Btu/lb in the nozzles. The nozzle angle is 20°. Calculate, 
for maximum work on the buckets, (a) the bucket speed, (6) the entry and exit angles 
of the buckets, (c) the final (residual) velocity of the fluid as it leaves the stage, ( d ) 
the force applied on the buckets per pound of fluid flow per second, and ( e ) the work 
per pound of flow. Neglect approach velocity to nozzles. 



























STEADY FLOW OF FLUIDS—THE TURBINE 


255 


Solution: 

(a) Referring to Fig. 11:11, V x = (223.8) (40)* = 1415 fps. 
For maximum work, 


Vb = \v I cos a = 


. 1415 cos 20° (1415) (0.94) 


2 2 

(6) V 2x = Vi cos a - Vb = Vi - - 1 


= 665 fps 


COS a — -$V\ COS a = -%V 1 COS a = 665 fps 


0i = 02 = tan 


-i Vl sin <* _ 1415 sin 20° _ (1415) (0,342) 


V 


2x 


665 


665 


= 0.728 


or 


0! = 02 = 36° 

(c) F 4 = Vi sin a = (1415) (0.342) = 485 fps 


(d) F = 


V 2x - Vz x 665 - (-665) 


= 41.3 lb per pound of flow per second 


g 32.2 

(e) W = FVb = (41.3) (665) = 27,400 ft-lb, or 35.3 Btu per pound of flow, 


Checking this result against the kinetic energy removed from the steam: 



2g 


1415 2 - 485 2 
64.4 


27,400 ft-lb 


11:11. Reaction Staging. Figure 11:14 diagrams a series of four pure 
reaction stages. Each stage consists of a row of stationary nozzles and a 
row of moving nozzle buckets. The entry blade angle to both is 90° in 
the pure reaction stage. (The real reaction stage usually employs a 
smaller entry angle in order to reduce the required number of stages.) 
Figure 11:15 shows a velocity diagram for one of these stages. It is 
observed that the purpose of the stationary nozzles is to give the fluid 
sufficient velocity so that its velocity relative to the bucket will be in a 
direction normal to the plane of bucket rotation; this permits the entry 
angle of 90° mentioned above. The velocity V 2 is the approach velocity 
to the moving nozzle, and the jet velocity is increased to F 3 and changed 
to a rearward direction by the action of that nozzle. In fact the moving 
nozzle duplicates the expansion of the stationary nozzle, each developing 
the same exit velocity at the expense of equal enthalpy drops. But the 
velocity F 3 is relative to the moving nozzle bucket and must be vectorially 
combined with Vb to give V 4 , the absolute velocity at exit from the stage 
(the approach velocity to the succeeding stage). The variation of pres¬ 
sure, enthalpy, and absolute velocity through the series of reaction stages 
is also included in Fig. 11:14. The pressure and enthalpy drops are con¬ 
tinuous through both stationary and moving blades; the curve of absolute 
velocity resembles that for impulse staging. 

Based on the proportions of Fig. 11:15 and following the method used 
in Art. 11:10, we may write for the reversible pure reaction stage 











256 


BASIC ENGINEERING THERMODYNAMICS 


V 2 X = v lx - V b = Vi cos a - V b 

Vz x = -Vi x = - Vi cos a _ 

Tir lt (V 2x - Vz x )V b (27i cos a - Vb)V h 
W per lb = - = --- 

g g 

dM _ 2 Fi cos « — 2 F& = q or y b = y l cos a (maximum work) 


dV b 


g 


W per lb = C0S2 a ~ C ° S2 °! = ^ C ° S2 a - = 2 7 cos 2 a 

max pei 1 U 0 g 2g 


As the nozzle angle approaches zero, the approach velocities to both 
stationary and moving nozzles approach zero and the maximum work of 
the stage approaches 2(7i 2 /2 g). But this is equivalent to the sum of the 



Fig. 11:14. Enthalpy, pressure, and velocity variation through a reaction turbine. 

kinetic energies generated in the two nozzles of the stage and therefore to 
the total stage drop of enthalpy. The reversible character of the process 
is not altered by giving the angle a a finite value; this would merely have 
the effect of providing an approach velocity to each row of nozzles, moving 
and stationary. 








































STEADY FLOW OF FLUIDS—THE TURBINE 


257 



Fig. 11:15. Velocity diagram—reaction stage (maximum work). 

Exam-pie 11:11. The total isentropic enthalpy drop across a reversible pure reac- 
tion stage is 40 Btu. The nozzle exit angles of both stationary and moving nozzles 
are 20°. Neglecting the approach velocity to the stationary nozzles, calculate, for 
maximum work in the stage, (a) the velocity at exit from the stationary nozzles, (6) the 
bucket speed, (c) the residual velocity of the fluid as it leaves the stage, ( d ) the force 
applied on the buckets per pound of fluid flow per second, and ( e ) the work per pound 
of flow. 

Solution: 

(a) Referring to Fig. 11:15, the exit velocity from the stationary nozzles is the 
approach velocity to the moving nozzles. Some of the kinetic energy developed in the 
stationary nozzles is therefore utilized in the moving nozzles. This has the effect of 
increasing the effective enthalpy drop to be divided between the moving and stationary 
nozzles above the specified 40 Btu for the stage. The enthalpy drop in the stationary 
nozzles may thus be estimated to approximate 21 Btu, and this would correspond 
to a velocity leaving the nozzle of (223.8) (21)^ = 1025 fps. The approach velocity 
to the moving nozzle would then be 1025 sin 20° = 350 fps, equivalent to an enthalpy 
drop of 350 2 /2 Jg = 2.46 Btu. The effective enthalpy drop per row of nozzles is 
therefore (40 + 2.46)/2 = 21.23 Btu. Thus Vi = (223.8) (21.23)* = 1030 fps. 
This differs negligibly from the velocity as estimated above and is the velocity at exit 
from the stationary nozzles. When the enthalpy drops in moving and stationary 
nozzles are equal, this is also the velocity V 3 of Fig. 11:15. To check, the approach 
velocity to the moving nozzles is Vi sin 20° = 352 fps, and the enthalpy drop in the 
moving nozzle 40 — 21.23 = 18.77 Btu/lb. Then, from Eq. (11:1), 

PV = 352 2 + (223.8)2(18.77) = 124,000 + 937,000 = 1,061,000 


or 


Fs = 1030 fps 

(5) y b = Vi cos 20° = V 3 cos 20° = (1030) (0.94) = 968 fps = V 3x 
(c) Vi = V 3 sin 20° = (1030) (0.342) = 352 fps 

Id) F = = ° ~| ~ = 30.1 lb/(lb) (sec) 

^ * 

(e) W = FV b = (30.1) (968) = 29,100 ft-lb, or 37.5 Btu/lb 

Note that the final kinetic energy in the fluid as it leaves the stage is 352 2 /2 Jg = 
2.5 Btu. When this is subtracted from the isentropic drop of enthalpy across the 
stage, the answer to part e is checked for this reversible stage. Points of interest 
which develop from a comparison of this solution with that of Example 11:10 are the 












258 


BASIC ENGINEERING THERMODYNAMICS 


higher bucket speed and yet lower leaving loss (residual velocity) characteristic of 
reaction staging as compared with impulse. 

11:12. Comparison of Impulse and Reaction Staging. The concept of 
the pure impulse and pure reaction stages has been based on the assump¬ 
tion of reversibility in flow; reversibility is not attainable in the flow of a 
real fluid, and the real-turbine stage can be classified as impulse or reac¬ 
tion only in a general sense. For example, some drop in pressure will 
always accompany the flow of a real fluid through the impulse bucket. 
However, a comparison of the idealized form of these two types of staging 
makes it possible to observe differences in design principles which cannot 
be disregarded in the design of the real turbine. It has been observed 
that close clearance between the moving bucket and the turbine casing 
need not be provided in the impulse stage because the drop in pressure 
across the bucket is small. This is not the case for the reaction stage 
since a major drop in pressure is maintained through the use of nozzles as 
buckets; leakage losses are of major importance in the design of the reac¬ 
tion stage. This explains why, in many turbines, impulse stages are used 
in the earlier, high-pressure stages of expansion, to be followed by reaction 
staging near the exhaust end of the turbine. The drop in pressure neces¬ 
sary to obtain a given isentropic drop of enthalpy is less in the low-pressure 
range; a reference to the Mollier diagram in the steam tables will confirm 
this fact for steam, but it is a common characteristic of all fluids. 

The drop in pressure across the buckets of the reaction stage makes it 
necessary that all buckets take part in the expansion simultaneously 
instead of only a section of the perimeter at one time, as for the impulse 
stage; otherwise the leakage of steam through idle-blade spaces could not 
be prevented. One of the results is that, in the effort to accommodate the 
greatly increased volume of fluid passing through the later stages of the 
reaction steam turbine, more dependence must be placed on increasing 
blade height and increasing rotor diameters. The governing of turbines, 
to operate at constant speed when the load is variable, is also affected. 
In the impulse turbine, loads lighter than that for which the turbine was 
designed may be carried by cutting off the steam flow through a part of 
the nozzles in* each row. This is called cutoff governing. This is not 
feasible as a means of governing the reaction turbine; instead, the throttle 
governor must be used, reducing the pressure of the steam as it is admitted 
to the reaction stage. This change from the pressure for which the nozzles 
were designed has the effect of lessening turbine efficiency. Although 
both impulse and reaction turbines are quite sensitive to changes in speed 
and load from those for which they were designed, the impulse turbine is 
somewhat the more flexible, on the basis of its ability to adapt itself to 
changing loads. 

An advantage in using reaction stages for the later stages of expansion 


STEADY FLOW OF FLUIDS—THE TURBINE 


259 


in the steam turbine derives from the increasing percentages of moisture 
in the steam as expansion proceeds to the lower pressures. This means 
that particles of dense water are carried in the steam jet as it issues from 
the nozzle and accounts for a tendency to erode the leading edge of the 
moving bucket. This effect is greatly reduced in the reaction stage, since 
the velocity of the entering jet is only that component which is normal to 
the plane of bucket rotation. Moreover, for the same total stage drop of 
enthalpy, the thermal energy of the steam is changed into kinetic energy 
in two steps in the reaction stage, and the maximum velocity reached is 
therefore only about 70 per cent as large as for the equivalent impulse 
stage. The desire to reduce erosion is the reason why steam is ordinarily 
supplied the turbine at high superheat; this reduces the percentage of 
moisture in the low-pressure stages. 

An examination of Figs. 11:10 and 11:11 indicates that the reversible 
impulse stage would produce no lengthwise (axial) thrust on the turbine 
shaft; the pressure is balanced on the two sides of the moving buckets, as 
are the axial components of fluid velocity at entrance and exit. Even in 
the case of the real impulse-staged turbine, the provision for counteracting 
thrust need only be minor. Reference to Figs. 11:14 and 11:15 will show 
that this is not the case for the reaction-staged turbine; the balancing of 
thrust is an important feature of the design of the reaction turbine. 

To utilize the same enthalpy drop at the same blade speed requires, in 
theory, the provision of twice as many stages of reaction blading as would 
be necessary if the blading were impulse. The expansion in a single row 
of stationary nozzles in the impulse turbine produces, in the limit, a 
velocity of 2F&, which is removed in the following row of buckets. The 
stationary nozzles of the reaction stage need develop a velocity of only V b , 
at the expense of an enthalpy drop only one-fourth as great as for the row 
of impulse nozzles. Although the buckets of the reaction stage also drop 
the enthalpy by the same amount, the total stage drop of enthalpy for the 
reaction stage is thus only half as large as that in the single expansion of 
an impulse stage having the same bucket velocity. The larger number 
of stages required is one of the principal factors in making reaction staging 
more expensive than impulse. 

11:13. Stage Efficiency. The efficiency of the reversible stage, like 
that of the reversible nozzle [see Eq. (11:14)] is 1. The efficiency of the 
real turbine stage is less than 1 only because of irreversibilities that have 
accompanied the flow of the fluid through the stage. The nature of these 
irreversibilities and their effect on flow through the nozzle have already 
been discussed. Let us examine the irreversibilities that are incurred in 
flow after the stationary nozzle has been passed. 

There will always be some drop in pressure across the bucket of a real 
stage, even if they are shaped like the (impulse-type) blades of Fig. 11 :7b. 


260 


BASIC ENGINEERING THERMODYNAMICS 


Thus the pressure at exit from the impulse stage will be at least slightly 
lower than the pressure at exit from the stationary nozzles. In Fig. 
11:16a the coordinates are those of the Mollier chart. The condition of 
the fluid at entrance to the stationary nozzles is denoted by the position 
of point 1 and that at exit from the real nozzle as point 2 '. The condition 
at exit from a reversible nozzle is located as point 2, lying on the same line 
of constant pressure as 2'. The small drop in pressure in the buckets of 
the impulse stage is the difference between p^ and pz) in the reversible 
stage, the work on the buckets per pound of fluid is hi — hz when this 
further drop in pressure is considered. Because of irreversibility in flow 



s 

(a) Impulse 



s 

(b) ReacHon 


Fig. 11:16. Nozzle-bucket and stage efficiencies. 


through the buckets, little of this additional isentropic enthalpy drop is 
realized in the form of work on the bucket, and the condition at exit from 
the bucket is represented by the position of point 3'. When Eq. (3:5) is 
applied to the adiabatic process, using stations at entrance to the station¬ 
ary nozzle and at exit from the moving bucket and neglecting differences 
in kinetic energy at these two points, the work performed on the buckets 
per pound of fluid flow may be shown to be hi — hy] the corresponding 
work for the reversible stage is hi — h 3 . The ratio of these enthalpy 
drops is the nozzle-bucket efficiency, or 


Vnb 


hi — hz f 
hi — hz 


(11:25) 


where the notation is that of Fig. 11:16. 

An additional irreversibility occurs outside of either nozzle or bucket 
and causes the so-called rotation losses. The rotating disk on which the 
buckets are anchored is surrounded by the fluid, and some power is con¬ 
sumed in keeping it moving against the resulting frictional drag. This 
loss tends to be higher for the impulse stage because some of the buckets 
are not receiving steam from nozzles and act merely as paddles in stirring 









STEADY FLOW OF FLUIDS—THE TURBINE 


201 


up the fluid. Thus paddle-wheel work is performed on the exiting fluid, 
and its temperature (or quality) is raised at constant pressure, as indi¬ 
cated by the line 3'-3" of Fig. 11:16. The increase of enthalpy as the 
result of rotation losses is hz» — hz>, and this is at the expense of the work 
performed on the buckets. Thus the final condition of the fluid as it 
leaves the stage is represented by the position of point 3" and the net 
stage work by the difference hi — hz». The stage efficiency is the ratio of 
this enthalpy drop to the isentropic drop hi — hz, or 


Vs = 


hi — hz" 
hi hz 


hi — hz r 
hi — hz 


hy — hz' 
hi — hz 



(11:26) 


where L is the rotation loss in Btu per pound of fluid passing the stage, and 
the notation is that of Fig. 11:16. 

The terms nozzle-bucket efficiency and stage efficiency have exactly the 
same meaning when applied to a reaction stage, as in Fig. 11:166. The 
only differences between the two diagrams are the smaller drops in pres¬ 
sure in the stationary nozzles and the larger drop in the buckets and the 
smaller rotation loss indicated for the reaction stage. 

11:14. Turbine efficiency is the ratio of the work delivered by the 
turbine per pound of fluid flowing through it to the work delivered by a 
reversible adiabatic turbine operating between the same inlet state and 
the same final pressure as the real turbine. The efficiency of a turbine 
consisting of only a single stage would be identical with the efficiency of 
that stage. 

In Fig. 11:17 is shown a condition curve for the series of stages that are 
included in a turbine. For simplicity, only four stages are shown; the 
number is often much larger. In fact, if steam is the fluid and there is a 
total isentropic enthalpy drop of 400 Btu/lb through the turbine, it has 
been shown in Art. 11:9 that at least eight impulse stages must be 
employed if the use of expanding nozzles is to be avoided. In Fig. 11:17, 
the lines labeled p 1 , p 2 , pz, etc., are lines of constant pressure on this Mollier 
chart and represent the pressures at entrance to the turbine and at exit 
from each of its stages, respectively. Their slopes increase with increas¬ 
ing temperature, according to Eq. (8:6), and they therefore diverge as 
shown in the figure. The state of the fluid at entrance to the first stage of 
the turbine is designated as point 1 and, at entrance to succeeding stages, 
as 2", 3", and 4". If the expansion through the turbine had been isen¬ 
tropic, the final state would have been as designated by the position of 
point 5, and the work of the ideal turbine is therefore hi — /i 5 . The work 
actually delivered by the turbine is the same as the sum of the work 
delivered by the individual stages, or (hi — h 2 ») T (hv r — hz» ) + (hz» 
— h v >) + (/i 4 " — hyi) = hi — hy>. Based on the definition of turbine 

efficiency, 






262 


BASIC ENGINEERING THERMODYNAMICS 


Vt = 


h i — hy 
hi — h$ 


(11:27) 


where rp is the turbine efficiency, a decimal fraction, and the notation is 
that of Fig. 11:17. 



Fig. 11:17. Turbine condition curve (data from Example 11:14). 


Point 2" in Fig. 11:17 corresponds to the state of the fluid at entrance 
to the second stage (or to point 3" of Fig. 11:16). From Eq. (11:26), the 
efficiency of the first stage, based on the notation of Fig. 11:17, is 

hi — / 12 " 

V ’ 1 = hi - h 2 

Similarly, the efficiencies of the succeeding stages are, respectively, 


VS2 


/ 12 " — h 3 " 
hy — hy 


Vs 3 


hy — hi" 
h%" — hi’ 


hi" — h 




5 " 


hi" — h 


5 ' 


and the average efficiency of all the stages may be written as 

hi — / 15 " 

^ Savg (hi — /12) “b (hy — h&) -f- (/i3" — hi') + (hi" — hw) 


(11:28) 


Because of the divergence of the lines of constant pressure on the chart, 














STEADY FLOW OF FLUIDS—-THE TURBINE 


263 


the denominator of Eq. (11:28) is larger than the denominator of Eq. 
(11:27). But the numerators are identical, and it is therefore evident 
that the efficiency of the multistaged turbine is greater than the average 
efficiency of its stages. 

Example 11:14. Steam enters a turbine that has four single-row impulse stages at 
100 psia, 500°F. The pressures at exit from the stages are 65, 40, 25, and 15 psia. 
The stage efficiencies are each 70 per cent. Neglecting the approach velocity to each 
stage, calculate ( a ) the temperature or quality at entrance to each stage, ( b ) the work 
per pound of steam expanded, and (c) the efficiency of the turbine. 

Solution: 

(a) With reference to the notation of Fig. 11:15 and employing the Mollier diagram 
for steam, hi = 1279; h 2 = 1237; h 3 = 1194; h A = 1156; h 6 = 1120; h 2 >> = 1279 
— 0.70(1279 — 1237) = 1250 Btu. This corresponds to a temperature of about 
435°F at entrance to the second stage and locates point 2". 

Returning to the chart and projecting vertically from point 2" to a pressure of 
40 psia, h z > is read as 1206 Btu. Then hy> = 1250 — 0.70(1250 - 1206) = 1219 Btu. 
At 40 psia, this enthalpy corresponds to a temperature of about 363°F. This is the 
condition of the steam at entrance to the third stage and locates point 3". 

Proceeding in a similar manner, is read from the chart as 1179 Btu and hi" = 
1219 — 0.70(1219 — 1179) = 1191 Btu. At 25 psia, this corresponds to a tempera¬ 
ture of about 301 °F and locates point 4". 

Also, hs’ is read from the chart as 1151 Btu and A 5 " = 1191 — 0.70(1191 — 1151) 
= 1163 Btu. This enthalpy, at 15 psia, corresponds to a temperature of about 238°F 
and locates point 5". 

W 

(b) y ~ hi — hy = 1279 — 1163 = 116 Btu per pound of steam 

( C ) Turbine efficiency = = 0.73 

11:15. The Axial-flow compressor is essentially a reaction turbine 
operated in reverse; it is sometimes an important part of the apparatus 
necessary to demonstrate the' cycle of the gas turbine, to be discussed in 
Chap. 12. It is staged to make it possible to compress gases such as air to 
high pressures without incurring serious irreversibility due to high veloc¬ 
ity. The blading is similar to the reaction blading shown in Fig. 11:14 
(though the angle relations are different in order that the blades may be 
better adapted for their new function), but the fluid enters at the right at 
low pressure and is discharged, at what would be the inlet to the turbine, 
at high pressure; the direction of bucket movement is opposite to that 
shown in the figure, and power is supplied from some external agency to 
drive the rotor. The moving buckets receive the fluid and increase its 
velocity, at the same time acting as diffusers in increasing the pressure; 
the stationary blades continue the diffuser action. The graphs of pres¬ 
sure, enthalpy, and velocity shown in the figure also apply to the axial-flow 
compressor but are traced from right to left. 

The condition line for the fluid passing through the axial-flow com- 




264 


BASIC ENGINEERING THERMODYNAMICS 


pressor is shown in Fig. 11:18. The compression is, ideally, isentropic, 
and the work ideally required for the compression of unit weight is h$ — hi. 
Because of irreversibility, the final condition of the fluid is at 5", and the 

work required for compression is 
hw — hi, greater than hb — hi. 
The ratio (hb — hi)/(hb" — hi) is 
called the adiabatic efficiency . 

The thermodynamic principle is 
the' same for the axial-flow com¬ 
pressor as for the adiabatic recipro¬ 
cating compressor, discussed in 
Chap. 9. This type of compressor 
is used when a continuous-flow 
type of compression is desirable and 
when its advantages of somewhat 
higher efficiency and lower frontal 
area (an advantage that is impor¬ 
tant when the compressor forms a 
part of an airplane power plant, be¬ 
cause of lowered air resistance) give 
it preference over the centrifugal 
compressor. The pressure ratio 
per stage approximates 1.2:1, and 
a relatively large number of stages 
are needed to obtain a given over-all pressure ratio; this means that it has 
the disadvantages of greater length and weight as compared with the 
centrifugal compressor. 



Fig. 11:18. 
compressor. 


Condition curve—axial-flow 


Problems 

1. A fluid flows through a horizontal channel. As it moves from an upstream 
section to a downstream section, its specific enthalpy decreases by 150 Btu/lb. If 
the velocity at the upstream section is 600 fps and the flow is adiabatic (though not 
frictionless), what is the velocity at the downstream section? 

2. A circular opening permits the passage of 2400 lb of steam per hour. If the 
velocity is 2000 fps and the steam has a density of 0.25 lb/ft 3 in the opening, what is 
its diameter in inches? 

3. As a fluid flows adiabatically through a horizontal channel, the velocity at an 
upstream section is 1000 fps. At a downstream section of the channel, the velocity 
is 2500 fps. What is the change of specific enthalpy between the two sections? 

4. An incompressible fluid is held in a tank. A horizontal nozzle is located in the 
wall of the tank at a level where the pressure inside the tank is steadily maintained at 
150 psia, and the liquid issues to the atmosphere through this nozzle. Assuming the 
flow to be isentropic, calculate the velocity of the issuing jet (a) if the density of the 
liquid is 62 lb/ft 3 and (6) if its density is 30 lb/ft 3 . 

5. Considering them to be perfect gases, calculate the critical pressure ratios for 
(a) carbon dioxide; (6) water vapor; (c) methane (CH 4 )[c„ = 0.45 Btu/(lb)(°F)]. 





STEADY FLOW OF FLUIDS—THE TURBINE 


265 


6. Air approaches a given nozzle with negligible velocity and at a pressure of 
100 psia. The pressure in the space to which the nozzle discharges is atmospheric. 
Will the mass rate of flow through the nozzle increase, decrease, or remain the same 
if the following individual changes are made in supply and exhaust pressures? (a) 
The discharge pressure is reduced to 5 psia. (6) The discharge pressure is raised to 
30 psia. (c) The discharge pressure becomes 70 psia. id) The supply pressure 
increases to 120 psia. ( e ) The supply pressure becomes 60 psia. 

7. Calculate the velocity of sound in the following perfect gases at a temperature 
of 70°F: (a) air; ( b ) water vapor; (c) carbon dioxide; ( d ) helium; (e) hydrogen; (/) 
methane (see Prob. 5). 

8. Air enters a reversible adiabatic nozzle at 100 psia, 200°F, and at low velocity. 
The discharge pressure is atmospheric. What is the temperature of the air at the 
throat of the nozzle? What is its velocity? Calculate the velocity of sound in air 
at the throat temperature, and compare. 

9. In a wind tunnel the model to be tested is held stationary in a moving stream 
of air, and the resulting forces and moments are measured. When the air is to be 
given high velocity, the nozzle principle may be used by allowing the air to pass 
through a constricted channel as the result of a pressure differential which is main¬ 
tained between its two ends. If the tests are to be made in the supersonic range 
(above the velocity of sound), discuss the necessary pressure differential and the 
location of the test section in the passage. 

10. Design a nozzle of round cross section to expand 3000 lb of air per hour from 
an initial pressure (at negligible velocity) of 140 psia and a temperature of 350°F to a 
final pressure of 15 psia. Assume frictionless adiabatic flow. 

11. Air enters a converging-diverging passage at 20 psia, 100°F, and with a velocity 
of 2000 fps. The area at the throat of the passage is 1 in. 2 ; the area at exit from 
the passage is large enough so that the velocity at that point may be considered 
negligible. Assuming the flow to be adiabatic and reversible, calculate (a) the 
temperature and pressure of the air at the exit and (6) at the throat. 

12. Assuming that the flow is adiabatic and reversible, find the required throat 
and exit areas of a nozzle to expand 3000 lb of steam per hour from 140 psia, quality 
of 0.98, to a final pressure of 15 psia. The entrance velocity is negligible. 

13. In Prob. 12, change entry conditions to 140 psia and 400°F, and design the 
nozzle. 

14. Redesign the exit areas of the nozzles of Probs. 10, 12, and 13 on the basis of 
an expected nozzle efficiency of 0.88. In each case, discuss the change in the proper¬ 
ties of the fluid as it leaves the nozzle. What is the velocity coefficient? 

15. Assuming a discharge coefficient of 0.99 will apply, determine the required 
throat areas in Probs. 10, 12, and 13. 

16. In Prob. 13, assume that the steam was supersaturated as it passed the throat 
of the nozzle, and calculate the coefficient of discharge for frictionless flow. At what 
maximum temperature could the steam have been supplied to give a coefficient of 
discharge greater than 1? 

17. Work Example 11:10 with the following changes in data: isentropic enthalpy 
drop = 100 Btu/lb; nozzle angle = 18°. Draw, to scale, a velocity diagram of 
the stage. 

18. Steam is supplied a reversible single-row impulse stage at 140 psia, 400°F, and 
at negligible velocity. The pressure at exhaust from the stage is 15 psia. The 
nozzle angle is 20°. Calculate, for maximum work on the buckets, (a) the bucket 
speed, (5) the entry and exit angles of the buckets, (c) the final velocity of the steam 
as it leaves the stage, (d) the force applied on the buckets per pound of steam flow 
per second, and ( e ) the work per pound of flow. 


266 


BASIC ENGINEERING THERMODYNAMICS 


19. The data are the same as for Prob. 18 except that the stage is a two-row reversi¬ 
ble impulse stage. The nozzle angle is 20° at entrance to the first row of buckets. 
Find, for maximum work per pound of steam, (a) the bucket speed; (b) the angle at 
which the steam is directed on the second row of buckets, relative to their plane of 
rotation; (c) the entrance and exit angles of both rows of buckets; ( d ) the force 
applied on the first row of buckets per pound of steam per second and on the second 
row; (e) the work per pound of flow delivered to each row of buckets. Draw, to 
scale, a velocity diagram of the stage. 

20. A reversible single-row impulse stage is designed to give maximum work when 

the bucket speed is 900 fps. The nozzle angle is 20°. (a) What is the enthalpy 

drop across the nozzles if the approach velocity is negligible? (6) What is the net 
enthalpy drop for the stage when effect is given to the residual velocity of the steam 
as it leaves the stage? 

21. The same as Prob. 20, except that the stage is a two-row reversible impulse 
stage. A three-row stage. 

22. If the velocity of sound in steam approximates 1700 fps at the state which 
exists in a reversible pure reaction stage having nozzles of the form shown in Fig. 11:14, 
what is the approximate maximum total enthalpy drop for which the stage maybe 
designed? 

23. Work Example 11:11 with the following changes in data: isentropic enthalpy 
drop across the stage = 60 Btu/lb; nozzle exit angles = 25°. 

24. Steam enters a reversible pure reaction stage at 10 psia, quality of 0.98, and at 
negligible velocity. The nozzle exit angles are 25°, and the bucket speed is 800 fps. 
The stage work is a maximum. What is the condition (pressure, quality, and 
velocity) of the steam at exit from the stage? What work is performed in the stage 
per pound of steam? If the steam leaving the stage is slowed reversibly to negligible 
velocity as it approaches the following stage, what is its state as it enters that stage? 

25. Steam is admitted to a turbine at 400 psia, 600°F. Condenser pressure is 
1 psia. Assuming zero nozzle angles and a bucket speed of 500 fps for all stages, 
calculate the minimum number of stages required for a reversible turbine if the stages 
are all (a) impulse single-row; (6) impulse two-row; (c) impulse three-row; ( d ) reaction. 

26. Steam is supplied a single-row impulse stage at 140 psia, 400°F, and at negligible 
velocity. The pressure at exhaust from the stage is 15 psia. The nozzle-bucket 
efficiency is 0.60, and the stage efficiency is 0.57. What is the rotation loss in the 
stage in Btu per pound of steam? What is the state of the steam as it leaves the 
stage? 

27. The enthalpy drop across a stage is 100 Btu per pound of steam. The stage 
efficiency is 0.55, and the rotation loss is 4 Btu per pound of steam. What is the 
nozzle-bucket efficiency? 

28. Steam enters a turbine that has three single-row impulse stages at 140 psia, 
400°F, and at negligible velocity. The pressure at the turbine exhaust is 15 psia. 
The turbine efficiency is 0.65. If equal work is performed in each stage and the 
stage efficiencies are equal, find the pressures at entrance to the second and the third 
stages. What are the stage efficiencies? 

29. Steam enters a turbine that has four single-row impulse stages at 140 psia, 
400°F, and at negligible velocity. The pressure at the turbine exhaust is 15 psia. 
The first stage has an efficiency of 0.58, the second 0.60, the third 0.62, and the 
fourth 0.64. Equal work is performed in all stages. By trial and error, locate the 
pressures at entrance to the second, third, and fourth stages, and calculate the turbine 
efficiency. Plot the condition curve on the Mollier diagram. (Note: A method of 
attack on this problem consists in assuming a reasonable and consistent turbine effi- 


STEADY FLOW OF FLUIDS—THE TURBINE 


267 


ciency, locating the state at exit from the turbine on the basis of this assumption, 
and checking the assumption against the given data.) 

30. Air enters an axial-flow compressor at 14.7 psia, 70°F, and is compressed to 
140 psia. The adiabatic efficiency of the compression is 0.80. At what temperature 
does the air leave the compressor? How much work is done on each pound of air? 
Approximately how many stages would be needed? 

Symbols 

A area 

c p specific heat at constant pressure 
C Vn coefficient of velocity of a nozzle 
Cd coefficient of discharge 
F force 

g acceleration of gravity 
h enthalpy of unit mass 
J proportionality factor 
' k ratio of the specific heats 
L rotation loss 
M mass rate of flow 
n a constant exponent 
p pressure, psi 

P pressure, psf; pressure in general 
Q rate of heat flow per-unit mass rate of flow 
R gas constant 
s entropy of unit mass 
t scalar temperature 
T absolute temperature 
u internal energy of unit mass 
v specific volume 
V velocity 

W rate of work delivery per unit mass rate of flow 
z elevation 

Greek Letters 

a nozzle angle to plane of rotation of the bucket 
/3 bucket angle to plane of rotation 
rj n nozzle efficiency 
rj nb nozzle-bucket efficiency 
77s stage efficiency 
1 ) t turbine efficiency 

Subscripts 

av average 
b bucket 
i ideal 

max maximum 
n nozzle 
rib nozzle bucket 
p constant pressure 
s constant entropy; also, stage 
t turbine 
v constant volume 


CHAPTER 12 


POWER—GAS SYSTEMS 

12:1. Introduction. For the production of power (the continuous 
manufacture of work) the use of some sort of cycle is indicated. In 
developing the basic concepts of thermodynamics, the heat-engine cycle 
has been used; it will be remembered that only heat and work may pass 
the boundaries of the heat engine. The Carnot and Stirling engines, 
introduced and described in Chap. 4, are examples of the heat engine, 
and both are typically associated with the gas system. The Carnot 
engine, for reasons discussed earlier, is a purely theoretical device and is 
not built in the form of a practical engine. The Stirling, on the other 
hand, in spite of the complexities and irreversibilities introduced by 
regeneration, was one of the first real engines to be based on the use of a 
gas system; its cycle was proposed by Robert Stirling in 1827. 

Heat enters both the Carnot and Stirling engines through the head of 
the cylinder; this cylinder therefore adds the functions of a heat exchanger 
to its duties in connection with the development of power. In Chap. 6 it 
has been brought out that, for maximum efficiency, heat should be received 
at as high an average temperature as is feasible and should be rejected at 
the lowest practical average temperature. The engineer normally pro¬ 
vides the heat supply by a process which involves the combustion of fuels; 
by means of this process, the temperature of the source can be maintained 
at 3000 to 4000°F. But this heat can be induced to enter the cylinder 
only by maintaining a differential of temperature between the outside and 
the inside of the cylinder. The highest temperature that can be attained 
by the working substance is limited by the metallurgical limit of the mate¬ 
rials used in the construction of the real cylinder. This is the highest 
temperature at which the walls can be trusted to retain a strength con¬ 
sistent with their other duties, such as the confinement of a fluid under 
pressure; it is about 1000°F for the metals ordinarily used by the engineer. 
Thus the effective source temperature (the maximum temperature attained 
by the enclosed system) must be below this metallurgical limit, and the 
attainment of high efficiencies is correspondingly handicapped. 

To avoid the limitation placed on the effective source temperature by 
factors such as the metallurgical limit, most gas-system power cycles 
employ internal combustion. Combustion takes place, in other words, 
within the engine. To prevent the walls from attaining too high a tern- 

268 


PO WER--GA S S YS TEMS 


269 


perature, a cooling fluid (usually water or air) flows around the cylinder. 
Here the differential of temperature is in the opposite direction, and, with 
eilective cooling, the walls may be prevented from reaching a level of 
temperature above that at which they may properly carry out their other 
functions. 

12:2. The Air-standard Cycle. The process of combustion is, in 
practice, an irreversible process, and, to make continuous operation of the 
internal-combustion engine possible, fresh supplies of fuel and air must be 
introduced, either intermittently or continuously, during the operation of 
the engine. Thus the internal-combustion engine not only receives heat 
and work across its boundaries but also accepts matter in the form of fuel 
and air and rejects it in the form of the products of combustion; it is there¬ 
fore not a heat engine. 1 

The internal-combustion engine does, however, operate on a cycle. 
This cycle may be read as the path traced on an indicator diagram taken 
during the operation of the engine and may conveniently be termed a 
cycle of the machine. The process of combustion will be represented on 
this diagram by a line, which may be a constant-volume, constant-pres¬ 
sure, constant-temperature, or some other sort of process which has been 
followed as the result of combustion, according to the conditions under 
which that combustion took place. Essentially, the same line could have 
been traced if heat had been supplied a closed system consisting of air 
alone, if the conditions that existed during the supply of that heat had 
been similarly controlled. Note that, in that case, no change in weight 
or in chemical composition need have taken place and the system could be 
handled as if composed of a pure substance. Similarly, the rejection of 
heat may substitute for the expulsion of the products of combustion, since 
it is no longer necessary to replace the charge. These two substitutions 
are introduced into what is called the air-standard cycle. 

The concept of the air-standard cycle makes it possible to change the 
machine cycle of the internal-combustion engine into an equivalent heat- 
engine cycle. It is our purpose to show the effects of various changes in 
the conformation of the cycle on the operating characteristics and the 
efficiency of the internal-combustion engine. This would be very diffi¬ 
cult, if we did not make use of this new concept, because of the large num- 

1 The process of combustion is, within certain limits, reversible. At extremely high 
temperatures, the products of combustion, formed at lower temperature, tend to 
dissociate into the reactants. If advantage may ultimately be taken of this behavior, 
much higher efficiencies may be hoped for than characterize the present-day heat 
engine. This possibility will receive more detailed attention in Chap. 19. 

Also, although the internal-combustion engine alone does not operate on a closed 
cycle, the cycle may be imagined as closed by including the effect of the processes of 
nature, which acts on the products of combustion to return them eventually to the 
form of the reactants. 


270 


BASIC ENGINEERING THERMODYNAMICS 


ber of interacting factors, such as changes in composition of the charge 
and its variable weight throughout the cycle, variable specific heats, and 
the effects of dissociation, that would necessarily be considered. The 
efficiency of the air-standard cycle, on the other hand, may be quite 
easily calculated in advance of the construction of the engine and indicates 
the maximum conceivable efficiency of an engine operating on that cycle. 
Although the real engine cannot attain so high an efficiency as that of the 
air-standard cycle, the ratios of the efficiencies of two air-standard cycles 
will serve as a valuable guide in the comparison of the two cycles. Also, 
the ratio of the actual efficiency of the real engine to the efficiency of its 
equivalent air-standard cycle is of some assistance in estimating the ratio 
of actual to perfect performance for that engine. This ratio will usually 
be about one-half to two-thirds; if it is much less than the lower figure, it is 
possible that improvement in design or operating conditions may result 
in worth-while gains. Of course, in the more advanced design stage, the 
factors ignored in the concept of the air-standard cycle must be taken into 
account. 

Other purposes are served as well by an analysis of the air-standard 
cycle as by the more complicated and laborious examination of the cycle 
of the real internal-combustion engine. For example, the general effect 
of changes in the precombustion treatment of the charge, and of changing 
the amount of fuel burned per cycle, may be shown. Also, a thermo¬ 
dynamic study of the air-standard cycle may suggest changes in the cycle 
itself that would be advantageous if they can be applied to the real engine 
and so may lead to an entirely new, and more satisfactory, cycle. 

12:3. The Otto Engine. The operation of the internal-combustion 
engine in most common present-day use is based on the cycle proposed in 
1862 by Beau de Rochas. An engine to operate on this cycle was first 
built in 1876 by Otto, a German engineer, and both cycle and engine are 
now known by his name. A schematic diagram, showing some of the 
essential features of the Otto engine, is presented in Fig. 12:1. Its 
idealized cycle of operation, as suggested by Beau de Rochas, is shown in 
Fig. 12:2. The engine and its cycle here illustrated are of the four-stroke- 
cycle type in which four strokes of the piston (two revolutions of the 
engine shaft) are required to complete the cycle; two-stroke-cycle opera¬ 
tion is also made possible by certain changes in the position of the valves 
and their method of operation and in the method of introducing the fuel 
mixture into the cylinder. Thermodynamically, as will be explained 
below, the four-stroke cycle becomes a two-stroke cycle when the concept 
of the air-standard cycle is introduced. 

Referring to Fig. 12:1, the downward motion of the piston on the suc¬ 
tion stroke, with the intake valve open, draws a charge of air through the 
intake line. On its way to the cylinder, fuel is drawn into this stream of 


POWER—GAS SYSTEMS 


271 


air in the carburetor, or mixing valve. This fuel may be either a volatile 
liquid, such as gasoline, or a fuel gas, such as natural gas; if the former, a 
carburetor is employed and, if the latter, a simple mixing valve. During 
the remainder of its passage to the cylinder, the fuel and air become thor¬ 
oughly mixed. 

At the proper point in the cycle that takes place within the cylinder, 
as described below, a spark is introduced into this mixture by an electrical 
system consisting of a battery and coil, or magneto, a timer, or distributor, 
and a spark plug set into the top of the cylinder. Since all of the fuel and 



Fig. 12:1. Internal-combustion engine— Fig. 12:2. Otto cycle. 

Otto cycle. 


all of the air necessary to its combustion are present, an explosion takes 
place. The final step in the cycle is the rejection of the burned gases to 
the atmosphere through the exhaust valve, preparatory to the entry of a 
fresh charge into the cylinder. The reciprocating motion of the piston is 
converted into rotary motion by means of the connecting rod and crank, 
and some sort of flywheel is necessary to store energy during the single 
power stroke and return it during the balance of the cycle. 

The water jacket shown here may be replaced by thin fins for direct air 
cooling. Cooling is necessary to protect the materials of which the cylin¬ 
der and piston are fabricated at the high temperatures that are developed 
within the cylinder; it is not necessary, or even desirable, in a strictly 
thermodynamic sense. The difficulty of properly cooling the piston 
accounts for the prevalence of the single-acting engine in internal- 
combustion design. The engine is governed by changing the position of 
the throttle valve at the top of the carburetor; this changes the weight of 








































272 


BASIC ENGINEERING THERMODYNAMICS 


fuel mixture taken into the cylinder without materially affecting the pro¬ 
portions of fuel to air. 

The detailed operations that take place within the cylinder are outlined 
below with reference to Fig. 12:2. 

1. A suction stroke , 5-1, during which a mixture of air and fuel is drawn 
into the cylinder at constant pressure; during this stroke the intake valve 
is open, the exhaust valve closed. 

2. Near the end of the suction stroke, the intake valve is closed 
mechanically, the piston reverses its direction of motion and, with both 
valves closed, carries out the compression stroke , 1-2. The mixture is com¬ 
pressed approximately adiabatically during this stroke, and the temper¬ 
ature of the charge rises with its pressure. 

3. Near the end of the compression the spark is introduced, exploding 
the charge while the piston is approximately stationary. The release of 
the heat of combustion thus takes place at nearly constant volume and 
raises the temperature and pressure, as indicated by the line 2-3 of Fig. 
12 : 2 . 

4. As the piston again changes its direction of motion, with both valves 
still closed, the expansion stroke , 3-4, provides for approximately adiabatic 
expansion of the products of combustion; during this stroke the tempera¬ 
ture gradually decreases. Near the end of the expansion stroke, the 
exhaust valve opens, and a part of the burned gases escapes from the cylin¬ 
der as the pressure drops along the line 4-1; this is a throttling operation 
but is not strictly steady-flow in character. 

5. The last stroke of the piston necessary to complete the cycle rejects 
the remainder of the burned gases at constant pressure as indicated by the 
line 1-5 of Fig. 12:2; this is called the exhaust stroke. Near the end of this 
stroke, the exhaust valve closes, and the intake valve opens, preparatory 
to retracing the cycle of operations described above. 

This cycle may be completed in two strokes of the piston instead of the 
four outlined above by providing for rejection of the burned gases and 
injection of a new charge at the end of the expansion and the beginning of 
the compression strokes. In real engines based on the Otto cycle, certain 
practical considerations, such as weight, size, cost, simplicity, and fuel 
economy, may cause preference to be given either the four-stroke or the 
two-stroke cycle. 

In converting the cycle described above to an equivalent air-standard 
cycle, the charge is air and air alone, and the same charge is used in suc¬ 
cessive traversals of the cycle. This makes the suction and exhaust 
strokes, 5-1 and 1-5, unnecessary, and the air-standard cycle becomes a 
two-stroke cycle. Also, the air-standard engine would, like the Carnot 
engine, require no valves. The combustion process 2-3 is replaced by the 
addition of heat in sufficient amount to bring about this rise in pressure at 


POWER—GAS SYSTEMS 


273 


constant volume, and the pressure drop 4-1 is accomplished by the extrac¬ 
tion of heat in appropriate amount. Furthermore, these processes are 
conceived as reversible processes, thus requiring that the concepts of a 
variable-temperature source and a variable-temperature refrigerator be 
introduced. The compression and expansion strokes, 1-2 and 3-4, are 
idealized as reversible and adiabatic in the Otto air-standard cycle. Not¬ 
ing that all processes of which it is comprised are individually reversible, 
we observe that the Otto air-standard cycle is a reversible cycle. How¬ 
ever, its reversibility does not., com¬ 
pare with that of the Carnot and 
Stirling cycles, which were designed 
to receive heat from a source at 
constant temperature and to reject 
heat to a refrigerator at some lower, 
but constant, temperature. 

To make the comparison between 
the air-standard Otto and the Carnot 
cycles more vivid, let us place them 
on the same TS diagram, as shown in 
Fig. 12:3. On this figure, the Otto 
air-standard cycle is shown as 1-2-3-4. 

This is a reversible cycle, and a com¬ 
parative idea of its efficiency may be 
obtained by comparing the area en¬ 
closed within the cycle with the area 
6-2-3-5 under the line 2-3, represent¬ 
ing the heat-supply process. 1 But 
there is available to the engineer, 
through the process of combustion, 

what amounts to a constant-temperature source at between 3000 and 
4000°F. This means that the temperature at point 3 may lie in this range. 
The basic refrigerator is the atmosphere, also characterized by a constant 
temperature, and it is evident that the lowest temperature of the cycle (at 
point 1) cannot fall below that of the atmosphere. If we conceive a 
Carnot engine operating between these constant temperatures of source 
and refrigerator, its cycle would be traced as l-2c-3-4c of Fig. 12:3 and its 

efficiency is ar6a r S Q - • The advantage of the Carnot in respect to 
area o—^c-o-o 

its efficiency is evident; its disadvantage is that its performance cannot be 
translated into terms of the operation of a real engine, while real Otto 
engines may be built and operated. 

Before leaving Fig. 12:3, let us note that, as the temperature rise during 

1 For the basis of this statement, the reader is referred to Art. 6:5. 



Fig. 12:3. Comparison of Otto 
standard and Carnot cycles. 


air- 













274 


BASIC ENGINEERING THERMODYNAMICS 


combustion decreases, point 3 will move downward along the curve 3-2 
toward point 2. Let us assume that this decrease in the heat suppty to 
the air-standard cycle has been carried to the point where point 3 is onty 
infinitesimally to the right of and above point 2. The result is shown in 
the small cycle which is pictured as lying alongside the isentropic com¬ 
pression line 1-2; it is evident that, as the heat supply decreases, the 
efficiencies of the Otto air-standard and of the Carnot cycles approach 
each other. This is, however, of little importance in the application of 
the cycle as the basis of the operation of a k real engine. It will be noted 
that the small Carnot cycle has an efficiency much less than that of the 
large Carnot cycle, which, in turn, was based on the available tempera¬ 
tures at which heat could be supplied and rejected. The thought may 
occur to us that, by carrying the isentropic compression to, or nearly to, 
point 2-c and reducing the heat supply, an improvement in performance 
of the Otto engine may be secured. But, in the real engine, line 1-2 repre¬ 
sents the compression of the charge, which is a mixture of fuel and air, 
preliminary to combustion. Point 2 may not be raised above the kindling 
temperature of the fuel, or a premature explosion will take place, dis¬ 
rupting the cycle. 

Let us proceed to a thermodynamic analysis of the Otto air-standard 
cycle. This is now a heat-engine cycle, and, from Eq. (2:15), its efficiency 


But 

and 


V = 


Qs Qr 
Q s 


(2:15) 


Q s = Mc v (T* - T 2 ) 
Qr = Mc v (T 4 — T i) 


where M is the mass of the air charge and the notation is that of Figs. 12:2 
and 12:3. 

Thus 


Qs — Qr _ Mc v (T 3 T 2 ) Mc v (T 4 T1) __ T 4 — T\ , 0 , 

v Qs Mc.(T, - Tt) ““ T-. - T, 

From Fig. 12:2, V 2 = V 3 , and V\ = V 4 , and, based on Eq. (9:40), 

T 1 _ (VX' 1 = (Vs\ k ~ l = T 4 

T 2 \ 7 i / \vj T 3 


which may be expressed in the form 










POWER—GAS SYSTEMS 275 


where c is a proportionality constant. In terms of this constant, Eq. 
(12:1) becomes 


„ = i _ Tt-Ti = _ cT i - T x _ _ (c - 1 )7\ 

Ti - T 2 cT 2 - T t (c - 1 )T 2 



( 12 : 2 ) 


T i and T 2 are the temperatures at the two ends of a reversible adiabatic 
(isentropic) process, and, substituting equivalent expressions for T\/T 2 
in terms of pressure and volume [see Eq. (9:40) or Table 9:2], the following 
may also be written: 


*7 = 1 
*7 = 1 



(12:3) 

(12:4) 


The ratio Fi/F 2 is called the compression ratio of the engine; in the Otto 
engine it is equal to the expansion ratio F4/F3. It will be noted that the 
efficiency of the Otto air-standard cycle depends solely on the compression 
ratio for this cycle and that it increases as the compression ratio increases. 
Thus the efficiencies of the large and the small Otto air-standard cycles of 
Fig. 12:3 are identical; for process 1-2 is the same for both cycles and 
represents the same compression ratio. In practice, the compression 
ratio is limited by the ignition characteristics of the fuel used, as has been 
explained above. Fuels must be gases, or liquids that can be vaporized 
for introduction into the air stream in the carburetor. Much progress has 
been made in the improvement of such fuels so that the ignition of the 
mixture will not take place until higher temperatures are attained, thus 
making the use of higher compression ratios, with correspondingly higher 
efficiencies, possible. This research in fuels has been directed by the 
result of analyses of the type carried out above and serves as an example 
of the manner in which thermodynamics may be applied to guide develop¬ 
ment. Originally, compression ratios of 5 were considered high; with the 
present selection of fuels, compression ratios above 8 may be used. The 
ability of a fuel to delay its ignition until higher temperatures are reached 
is expressed in terms of its octane number, higher octane numbers corre¬ 
sponding to higher kindling temperatures. Little or no gain in efficiency 
results when a fuel of higher octane number than is required to prevent 
premature ignition is used in an engine of low compression ratio; the 
octane number has no necessary relation to the heat of reaction of the 
fuel, but the higher octane fuels are usually more expensive. 

The ratio of the volume V 2 to the volume Vi — V 2 swept through by 
the piston (the piston displacement) is known as the clearance of the 
engine; thus 

F 2 


Clearance = 


Fi - F 2 


(12:5) 







276 


BASIC ENGINEERING THERMODYNAMICS 


in which the clearance is expressed as a decimal fraction. The clearance 
is directly dependent upon, and follows from, the compression ratio. 

The Otto-cycle engine cylinder combines the function of a compressor 
with that of an engine, the net output per cycle being the difference 
between the work performed on the piston during the expansion stroke 
and the (negative) work of compression. This net work is usually 
expressed in terms of the mean effective pressure (mep), which is the aver¬ 
age height of the pressure-volume diagram. The mep is an index to the 
size of cylinder required for a given net work output per cycle. The size 
is often a critical factor in the selection of a suitable type of prime mover. 

The mep of the air-standard cycle may be calculated by dividing its 
area, as plotted on pressure-volume coordinates, by the length of the 
cycle; or since the area of this reversible cycle is the net work of the cycle 
and is, in turn, equal to the net heat flow, 


mep 


W W 
V i - V 2 V D 


J(Qs - Q r ) 



( 12 : 6 ) 


where V D is the piston displacement. The terms indicated mean effective 
pressure (imep) and brake mean effective pressure (bmep) are both used in 
connection with the real engine. The imep is obtained by dividing the 
area of an actual indicator diagram by the length of the diagram, which 
amounts to dividing the net work performed on the piston by the piston 
displacement. In calculating the bmep, the net work delivered by the 
engine is divided by the piston displacement; the bmep is less than the 
imep because of the frictional losses in the engine. 

Example 12:3. An Otto air-standard cycle has a compression ratio of 7. The 
lowest and highest temperatures of the cycle are 90 and 3000°F, respectively. The 
pressure at the beginning of compression is 14 psia. Calculate (a) the pressures 
and temperatures at the key points of the cycle, (6) the heat supplied, the heat rejected, 
and the net work per pound of air per cycle, (c) the efficiency, (< d ) the clearance, and 
(e) the mep of the cycle. 

Solution: 


(a) From data: pi = 14 psia; T i = 550°R; T 3 = 3460°R. 

Tt = 7\ (k) 1 -' = (550) (7)°" = (550) (2.18) = 1200°R 
p 2 = pi (yY = (H)(7) 1 " = (14)(15.2) = 213 psia 

p3 = pl GO = 213 1555 = 615 psia 

T, = T, (TV * = 3460 GY ’ = 3460 (ols) = 1590°R 
P) = P3 (£)* = 615 (i)‘ J = 615 ( T V) = 40.5 psia 







POWER—GAS SYSTEMS 


277 


(b) Qs = Mc v (Tz — T 2 ) = 0.171(3460 — 1200) = 386 Btu/lb per cycle 
Qr = Mc v (T 4 — Ti) = 0.171(1590 — 550) = 178 Btu/lb per cycle 

h IJ = Qs Qr = 386 1/8 = 208 Btu/lb, or 162,000 ft—lb /lb per cycle 

(c) v 


1 /FoY" 1 , /1\ 0 - 4 

= =1 -(t) =1-0.46=0.54 


Checking, 

Qs — Qr _ 386 — 
77 ~ Qs ~ 386 


178 


= 0.54 


(d) ». - = 14.5 ftyib; v, = ^ = 2.07 ftVlb 

Fd = t>i - i> 2 = 14.5 - 2.07 = 12.43 ft 3 /lb 

Clearance = = 0.166, or 16.6 per cent 

, \ W 162,000 . 

W mep = V~ D - (12.43)(144 ) - 90 ' 5 psl 


12:4. The Diesel Cycle. In 1893 Dr. Rudolf Diesel, recognizing the 
desirability of supplying heat to the cycle of an internal-combustion 
engine at as high an average temperature as possible and the handicap 
incurred in this respect by the Otto engine, due to its compression of a 
combustible mixture, proposed an engine which would compress a charge 
consisting of air alone to high temperatures. When the compression 
stroke had been completed, the fuel was to be injected into the cylinder 
and to burn progressively as it came in contact with the charge of air; 
because of the high temperature of the compressed charge, no spark or 
other ignition system would be required. 

Dr. Diesel’s original idea was to carry the compression to a temperature 
as high as could be safely used within the cylinder and then to control the 
rate of fuel injection, as the piston started on its power stroke, so that the 
combustion would take place isothermally; a constant-temperature 
source, as in the Carnot engine, would thus be simulated. He also hoped 
to approach isothermal rejection of heat at a later point in the cycle by 
injecting water into the working fluid. Dr. Diesel soon found these 
isothermal processes to be impractical, but, from his proposal, the form of 
a present-day cycle that has come to be known as the diesel cycle has been 
derived. The diesel cycle, as we know it, retains his concept of a high 
compression of a charge consisting of air alone and the injection of the fuel 
into the cylinder at a controlled rate during the first part of the power 
stroke; the rate of fuel injection, however, is calculated to give constant- 
pressure, rather than constant-temperature, combustion. The other 
processes of the Otto cycle are retained. 

The diesel engine is diagramed in Fig. 12:4. In this engine, the charge 
drawn into the cylinder is air alone, and a carburetor is not used. No 
ignition system is required, since the temperature of the charge after com- 












278 


BASIC ENGINEERING THERMODYNAMICS 


pression is high enough to ensure ignition of the fuel as it enters the cylin¬ 
der; instead, a fuel-injection system is substituted, capable of raising the 
pressure of the fuel high enough so that the fuel not only can enter the 
cylinder against the pressure of the compressed charge but also will be 


thoroughly atomized as it enters. 



Fig. 12:4. Internal-combustion engine— 
diesel cycle. 


A much less restricted choice in the 
matter of the type of fuel to be 
used is available than for the Otto- 
cycle engine, since the fuel need not 
be volatile; cheaper fuels may 
therefore be utilized. 

The fuel-injection system con¬ 
sists essentially of a fuel-injection 
pump, driven from the engine shaft, 
and a fuel-injection nozzle, designed 
to atomize the fuel thoroughly as it 
enters the cylinder of the engine. 
The fuel burns progressively as it 
enters and comes in contact with 
the heated air in the cylinder. The 
rate of fuel delivery must be very 
carefully proportioned to the 
change of volume of the charge as 
the piston begins its power stroke if the pressure is to remain even approx¬ 
imately constant during the period of combustion. Governing is effected 
by designing the fuel-injection system so that the time period over which 
injection takes place can be varied either manually or automatically. 

The intake and exhaust valves and the valve-operating mechanism of 
the diesel engine may be quite similar to Otto-cycle engine practice. The 
compression ratio is no longer limited by the kindling temperature of the 
fuel, but only by the practical limits of pressure and temperature that the 
cylinder can withstand. Indeed, the limit on compression ratio now 
becomes a lower limit, since the temperature of the compressed charge of 
air is relied upon to ignite the fuel as it enters the cylinder. Compression 
ratios used in practice are about 15:1. 

When the operations comprising the cycle of a real diesel engine are 
idealized in the form of the equivalent air-standard cycle, they appear as 
illustrated on PV and TS coordinates in Fig. 12:5. With reference to this 
figure, the successive processes may be itemized as follows: 

1. The adiabatic compression of a charge of air to a high pressure and 
temperature, represented by the line 1-2. 

2. The supply of heat at a rate sufficient to maintain a constant pres¬ 
sure (equivalent to the progressive injection and burning of the fuel in the 
real engine) during the early part of the power stroke, represented by line 
2-3. At point 3 this heat supply ceases. 



































POWER—GAS SYSTEMS 279 


3. Isentropic expansion of the air charge to the end of the power stroke, 
3-4. 

4. Rejection of heat from the charge, 4-1, lowering the pressure at con¬ 
stant volume, and preparing the charge for a retraversal of the cycle. 



Fig. 12:5. Diesel 



(This is equivalent to the rejection of the charge of burned gases in the 
real engine.) 

The efficiency of the air-standard diesel cycle may be calculated as 
below: 


Q s = Mc p (Ts - T t ) Q r 
Qs - Qn _ Mc v {T 3 - Ti) - Mc v {T 4 
Qs " Mc p (Ts - T t ) 


Mc v (Ta — T i) 

TO 


T* - Ti 

HTi - TO 


(12:7) 


The temperatures T h T 2 , Tz, and T 4 for substitution in Eq. 12:7 may be 
found, for a specific cycle, by an application of the information tabulated 
in Table 9:2. The same definitions of compression ratio, clearance, piston 
displacement, and mep apply as for the Otto cycle. The expansion ratio 
Va/ Vs is smaller than the compression ratio and can be altered inde¬ 
pendently of the compression ratio by shortening or lengthening the time 
during which heat is supplied and thus moving point 3 to the left or right. 
The efficiency, as expressed in Eq. (12:7), is thus evidently not alone a 
function of the compression ratio for this cycle. The position of point 3, 
in terms of the cutoff percentage, is expressed as 

Cutoff, per cent = ^^ (100) 

v 1 — ’ 2 


( 12 : 8 ) 















280 


BASIC ENGINEERING THERMODYNAMICS 


The effect of a variation in the percentage of cutoff on the efficiency of 
the diesel air-standard cycle may be shown as follows: 


V 


= 1 


T 4 - Ti 

k(T 3 - T 2 ) 


= 1 - 


T,\ Ta/T 1 - 1 
Tj k(T,/T 2 - 1) 


(12:9) 


But 


T 2 


Vz 

V 2 


E 3 = EJ t 4 


-<tT 


Ti = T< 



k —1 


F 4 = Vi 


and 


T* 

Ti 


TzWz/Vi) 1 *- 1 _ VjVt/VX- 1 _ F./F.V 1 _ (Vz\ k 

t 2 (f 2 /f 1 )^- 1 f 2 \v 2 /vj v 2 \v 2 J \vj 


Substituting these values in Eq. (12:9), 

>*y _i (F 3 /F 2 ) fc 


V 


= 1 - 


- 1 


k(Vz/V 2 - 1) 


( 12 : 10 ) 


The ratio F 3 /F 2 is unity for zero percentage of cutoff and increases as the 
cutoff lengthens; it may be called the load ratio. The ratio V 2 /V\ is the 
reciprocal of the compression ratio. Examination of Eq. (12:10) indi¬ 
cates that the efficiency of the air-standard diesel cycle is a function of 
both the compression ratio and the percentage of cutoff, or load ratio. A 
comparison of Eq. (12:10) with Eq. (12:3) indicates that the expression 
for the efficiency of the air-standard diesel cycle closely resembles that 
which gives the efficiency of the Otto cycle, differing from it only because 
of the multiplication of the second term of Eq. (12:3) by the load-ratio 
(VJV 2 ) k - 1 

factor 7 /T7 /T v - r-r in Eq. (12:10). As the load ratio approaches unity 

K>\ y 3 / v 2 — i) 

(i.e., as the cutoff approaches zero), the load-ratio factor approaches 1 as 
a limit and the efficiency of the air-standard diesel equals that of the air- 
standard Otto cycle. As the load ratio increases (the cutoff lengthens), 
the load-ratio factor increases above 1 and the diesel-cycle efficiency is less 
than the efficiency of an Otto cycle of the same compression ratio. This 
comparison is not fair to the diesel cycle since, as a real engine, it can make 
use of much higher compression ratios than can the real Otto-cycle engine. 
A more equitable basis of comparison would be to assume that the maxi¬ 
mum temperatures reached in the two cycles were the same; as will be 
shown later, when this stipulation is made the diesel has the higher 
efficiency. 


Example 12:4. A diesel air-standard cycle has a compression ratio of 15. The 
lowest and highest temperatures of the cycle are 90 and 3000°F, respectively. The 
pressure at the beginning of compression is 14 psia. Calculate (a) the pressures and 
temperatures at the key points of the cycle, (b) the heat supplied, heat rejected, 
and net work per pound of air per cycle, (c) the efficiency, (d) the clearance in per cent. 
( e ) the mep of the cycle, and (/) the cutoff percentage. 












POWER—GAS SYSTEMS 


281 


Solution: 

(a) From data: p x = 14 psia; Ti = 550°R; Tz = 3460°R. 

Tz = T x = (550) (15) 0 - 4 = (550)(2.96) = 1630°R 

p 2 = pi = (14)(15) 1 - 4 = (14)(44.5) = 623 psia = p 3 

Vj = Tz = 3460 = Vt _ F 3 _ Fs V 2 _ 9 19 / 1 \ _ _ n .. 

V 2 T 2 1630 ' F 4 ~ F 2 Vi ~ 212 \ 15/ “ 0-141 

= ^ ( 7 ;) = (3460)(0.141) 0 - 4 = (3460) (0.457) = 1580°R 

Pa = p 3 = (623)(0.141) 14 = (623)(0.064) = 40 psia 

(b) Qs = Mc p (Tz — T 2 ) = 0.24(3460 — 1630) = 439 Btu/lb per cycle 
Qr = Mc v (Ti — Ti) = 0.171(1580 — 550) = 176 Btu/lb per cycle 

W/J = Qs — Qr = 439 — 176 = 263 Btu/lb, or 205,000 ft-lb/lb per cycle 


(c) =l _ T 4 - Ti 

; V k(T 3 - T 2 ) 1.4(3460 - 1630) 

Checking, 

Qs ~ Qr 439 - 176 


1580 - 550 


= 1 


0.402 = 0.598 


v = 


= 0.598 


.598 


Qs 439 

or, using Eq. (12:10), 

_ , ( iy 4 2.121 4 - 1 n qqq ( l-86\ nt 

V \ 15/ 1.4(2.12 - 1) 1 0-338 Vl. 57 ) “ ° -t 

Note that the efficiency of an Otto cycle of this compression ratio is 1 — (tS) 0 ' 4 
= 0.662. 

(d) v ! = 14.5 ft 3 /lb [Ex. 12:3]; v 2 = vi/15 = 14.5/15 = 0.966 ft 3 /lb 
Fd = Vl - v 2 = 14.5 - 0.966 = 13.53 ft 3 /lb 


Clearance = 
( e) mep = 


v 2 0.966 


Vd 13.53 
W 205,000 




V d (13.53)(144) 


= 0.0713, or 7.13 per cent 
= 105 psi 


(/) Vz = ^v 2 = (2.12) (0.966) = 2.145 
v 2 

^ A (vz - 02 ) (100) (2.145 - 0.966) (100) 0 ^ 

Cutoff, per cent = -- = - ro ——-- = 8.7 per cent 

V d lo.5o 


12:5. The Dual Cycle. Originally, the atomization of the fuel as it 
entered the cylinder of the diesel engine was assisted by introducing it in 
a stream of secondary air at high pressure; the expansion of this jet of air 
as it entered the cylinder caused the fuel jet to be shattered into minute 
particles and to spread through the combustion space. The tendency in 
diesel design has been toward the elimination of this secondary air supply, 
so that the fuel alone is injected into the cylinder and dependence is placed 
on a high pressure differential across the fuel-injection nozzle for proper 















282 


BASIC ENGINEERING THERMODYNAMICS 


atomization. This has lead to a variant of the diesel cycle, called the dual 
(for dual-combustion) cycle; it is illustrated in Fig. 12:6 on PV and TS 
coordinates. The real engine that operates on an approximation of this 
cycle differs little in its construction from the diesel of Fig. 12:4, except 
that the compression ratio is usually somewhat smaller. The cycle of 
operation is slightly changed by advancing the time at which fuel first 
enters the cylinder, and the first part of the fuel charge is introduced while 
the piston is practically stationary; as a result, the first part of the com- 




Fig. 12:6. Dual air-standard cycle. 

bustion takes place at nearly constant volume. The balance of the fuel 
is injected during the early part of the power stroke, as in the diesel cycle, 
and the rest of the cycle is the same as the diesel cycle. To distinguish 
them from the true diesel, engines based on the dual cycle are called com¬ 
pression-ignition (Cl) engines. They retain the advantages of the diesel 
in permitting a wider selection of fuels and higher compression ratio than 
are available to the real Otto-cycle engine. It may be shown that, on the 
basis of the same compression ratio , the efficiency of the air-standard dual 
cycle is less than that of the Otto, but greater than that of the diesel. On 
the more practical basis of comparing the efficiencies of these cycles when 
the maximum temperature reached as the result of combustion is the same 
for each, the diesel enjoys a slight advantage over the dual cycle, but the 
latter is superior to the Otto. It is suggested that the reader derive an 
expression for the efficiency of the air-standard dual cycle in terms of the 
temperatures at the key points around the cycle of Fig. 12:6 and examine 
that expression with a view to determining what factors control the 
efficiency. 










POWER—GAS SYSTEMS 


283 


Example 12.o. A dual-combustion air-standard cycle has a compression ratio of 10. 
1 he constant-pressure part of the combustion takes place at a pressure of 623 psia. 
The lowest and highest temperatures of the cycle are 90 and 3000°F, respectively. 
The pressure at the beginning of compression is 14 psia. Calculate ( a ) the pressures 
and temperatures at the key points of the cycle, ( b ) the heat supplied at constant 
volume and at constant pressure, the heat rejected, and the net work of the cycle per 
pound of air, (c) the efficiency, and (d) the mep of the cycle. 

Solution: 


(a) From data: p 4 = 14 psia; 7b = 550°R; p 3 = p 4 = 623 psia; 7b = 3460°R. 
T 2 = Ti 1 = (550) (10) 0 - 4 = (550)(2.51) = 1380°R 

?>2 = pi = (14)(10) 14 = (14)(25.0) = 350 psia 


7b = 7b ^ = 1380 (= 2460°R 
P 2 \350/ 

Z_ 4 = Ia = 3460 Ane _ Vj F_ 4 _ F 4 _ V 4 V 2 1.408 
F 3 


= 1 408 = — • — = _ 4 = U __ 
T 3 2460 ' V 2 ’ V 5 V l V 2 V j 


10 


= 0.1408 


7 7 6 = 7b = (3460)(0.1408) 0 - 4 = (3460)(0.457) = 1580°R 

Ph = Pi {vS = ( 623 )(°- 1408 ) 1 ' 4 = (623)(0.064) = 40 psia 

(b) ( Qs)v = Mc v (Tz - T 2 ) = 0.171(2460 - 1380) = 184.5 Btu/lb 
(Qs) P = Mc p (T 4 - T 3 ) = 0.240(3460 - 2460) = 240.0 Btu/lb. 

Qs = 424.5 Btu/lb 

Qr = Mc v {T b - Ti) = 0.171(1580 - 550) = 176 Btu/lb 
W/J = Q s - Q r = 424.5 - 176 = 248.5 Btu/lb, or 193,000 ft-lb/lb 

r .x Qs - Qr 248.5 A ror 
(C) " = = 42475 = °- 585 

(d) vi = 14.5 ft 3 /lb; v 2 = 14.5/10 = 1.45 ft 3 /lb 

193,000 


mep = 


(14.5 - 1.45) (144) 


= 102.5 psi 


12:6. Comparison of Otto-, Diesel-, and Dual-cycle Efficiencies. 

Some attention has been given in preceding paragraphs to the comparison 
of the efficiencies of the Otto, diesel, and dual air-standard cycles. In 
Chap. 6 it has been pointed out that an excellent tool for the purpose is the 
temperature-entropy diagram. The comparison of these cycles on that 
diagram is shown in Fig. 12:7, the basis of the comparison being equality 
of the highest and lowest temperatures of the cycles (at points 3 o, 3 D , 4 d , 
and point 1, respectively). The outline of the Otto cycle is shown in 
solid, of the diesel cycle in dashed, and of the dual cycle in dash-dot lines, 
and the notation corresponds to that of Figs. 12:2, 12:3, 12:5, and 12:6. 
The subscripts 0, D, and d refer, respectively, to the Otto, the diesel, and 
the dual cycles. It will be noted that the cycles differ only at their tops, 
the isentropic expansion lines (3-4 for the Otto and diesel and 4-5 for the 
dual) and the lines representing the constant-volume rejection of heat 









284 


BASIC ENGINEERING THERMODYNAMICS 


(4-1 for the Otto and diesel and 5-1 for the dual) being the same for each 
cycle. The compression ratio is represented by the length of the line 1-2 
in each cycle and, based upon the premise adopted above, is greatest for 
the diesel, least for the Otto cycle. 

The efficiencies of these reversible cycles may be visually estimated as 
the ratio of their enclosed areas (their net work) to the area between their 
top lines (2-3 for the Otto and diesel, 2-3-4 for the dual) and the S axis, 

this area representing the heat sup¬ 
plied by the source; the advantage of 
the diesel and dual cycles over the 
Otto cycle is readily apparent. It is 
suggested that the reader make the 
same comparison, based on the same 
premise, between the cycles as they 
appear on PV coordinates and note 
that the same conclusions are not so 
evident. Also,’he may compare them 
on TS coordinates, but based on the 
assumption of equal compression 
ratios for each cycle, and check the 
results of his comparison against the 
equivalent, but more laborious, com¬ 
parison as made in Arts. 12:4 and 12:5. 
The purpose is, of course, to emphasize 
the advantage in the use of the TS 
diagram as a tool for comparing the 
efficiencies of reversible cycles. 

12:7. The Lenoir engine is of no present-day practical importance. It 
does have interest for us as the first form in which the internal-combustion 
engine was built and because it illustrates the way in which a knowledge 
of thermodynamic principles may guide research and lead to eventual 
improvement in heat-work apparatus. 

The Lenoir engine was a two-stroke-cycle engine, with intake and 
exhaust valves in the head of the cylinder. As the piston moved away 
from the head, with the intake valve open and the exhaust valve closed, a 
charge consisting of a mixture of fuel and air was drawn into the cylinder. 
Part way in the stroke of the piston, the intake valve was closed, and the 
charge then within the cylinder was ignited and exploded without previous 
compression. Because of this explosion, the pressure within the cylinder 
was raised, and, during the latter part of its stroke, work was performed 
on the piston as the pressure gradually decreased. At the end of this 
piston stroke the exhaust valve opened, and the burned gases were expelled 
during the exhaust stroke of the piston. At the end of this exhaust 



Fig. 12:7. Comparison of Otto, diesel, 
and dual air-standard cycles. 








POWER—GAS SYSTEMS 


285 


stroke, the exhaust valve closed, and the intake valve opened, preparatory 
to retracing the cycle. The cycle, idealized in the form of an approxi¬ 
mately equivalent air-standard cycle, is shown in Fig. 12:8. The actual 
cycle differed widely from this idealization, both because the rapid move¬ 
ment of the piston at the time of explosion made the combustion line 2-3 
vary greatly from the vertical rise of pressure pictured in the figure and 
because the pressure at the end of the power stroke did not necessarily 
correspond to the pressure at which the charge was inducted. 

In devising what is now known as the Otto cycle, Beau de Rochas was 
able to make a long step in the advancement of internal-combustion- 
engine theory because of his recognition of the thermodynamic principle 




Fig. 12:8. Lenoir air-standard cycle. 

that, for high efficiency, heat should be supplied the cycle at as high an 
average temperature as is practicable. The purpose of the precompres¬ 
sion of the charge, which he first introduced, was to raise this average 
temperature, and, since his day, precompression of the charge has been 
accepted as a basic principle to be applied in internal-combustion-engine 
design. It is suggested that the reader both derive an expression for the 
efficiency of the Lenoir air-standard cycle of Fig. 12:8 and make a com¬ 
parison on TS coordinates, in the manner of Fig. 12:7, with cycles that 
employ precompression. 

12:8. The Brayton (Gas-turbine) Cycle. If the adiabatic expansion 
line 3-4 of the PV diagram of Fig. 12:5 is prolonged until the pressure at 
point 1 of the cycle is reached, it is evident that an extra dividend of work 
will be obtained, equal to the toe-shaped area which is added when the end 
of this extended process is connected to point 1 of the cycle by a constant- 
pressure line. Moreover, this extra work of the cycle does not come at 
the expense of any increase in heat supplied the cycle, since the heat- 










286 


BASIC ENGINEERING THERMODYNAMICS 


supply line 2-3 need not change its position nor its length. The resulting 
cycle is known as the Bray ton air-standard cycle and is the basic cycle 
underlying the operation of the gas-turbine power plant. The Brayton 
air-standard cycle is shown on PV and TS coordinates in Fig. 12:9. The 
improvement in efficiency over the diesel is shown even more clearly by 
comparing the TS diagrams of Figs. 12:5 and 12:9. This comparison 




Fig. 12:9. Brayton air-standard cycle. 



Fig. 12:10. Schematic diagram of gas-turbine power plant. 


shows that the increase in net work results from a reduction in the amount 
of heat discharged to the refrigerator. 

The Otto, diesel, and compression-ignition power plants are typically 
built in the form of reciprocating engines, but the real power plant that is 
based on the Brayton cycle is a steady-flow device consisting of an air 
compressor and a gas turbine in series. Fuel is burned in the air as it 
leaves the compressor and still further increases its temperature and thus 
its volume as it enters the gas turbine for reexpansion to atmospheric 
pressure. A schematic diagram of the elements of the gas-turbine power 
plant is presented in Fig. 12:10. It is evident that the work required to 



















































































POWER—GAS SYSTEMS 


287 


drive the compressor is a charge against, and must be deducted from, the 
work delivered by the turbine in arriving at the net work of the entire 
plant. In Fig. 12:11, the work expended 
on compression is proportional to the 
area 1-2 -b-a, while the work of the 
turbine is represented by the area 
6-3-4-a. The net work of the cycle is 
therefore proportional to the area 
1-2-3-4, which is, of course, the Brayton 
cycle. Since the gas-turbine power 
plant is a steady-flow device, the vol¬ 
umes as shown in this figure would con¬ 
veniently be based on a unit mass of the 
fluid. In converting to the air-standard 
basis for convenient thermodynamic 
analysis, the only change that is neces¬ 
sary is to substitute a simple addition of 
heat sufficient to produce the change of 
temperature and specific volume between 
points 2 and 3 and a rejection of enough heat, between points 4 and 1, to 
change an air system of unit mass between those states. 

As an air-standard cycle, the Brayton cycle may be analyzed, as below: 

Qs — c p (T s — T 2 ) Qr = c p (T 4 — T \) 

Qs - Qr Cp(Ts - T 2 ) - c p (T 4 - 7\) 1 T 4 — Ti 

7 ? = —~- = l - 



Fig. 12:11. Work areas—Brayton 
cycle. 


Tt 

T x 

Then 



Qs 

Cfc—l)/fc 



or 


T 4 - T 1 


c p (T 3 — T 2 ) 

(A ;—\)/k rrt 

i 3 

” T 4 


cTi - T] 


Ts - T 2 

T 4 = T Z = 

T 1 T 2 ° 

(proportionality const) 


(C - l)? 1 ! 


77 = 1 - ^^ = 1 - ^^ = 1 - ^= 1 


T* - T 5 


cT 2 — 


(c - 1 )T 2 


Ti 

T 2 


( 12 : 11 ) 


Comparing this expression with Eq. (12:2), the efficiency of the air- 
standard Brayton cycle is found to be identical with that of the air- 
standard Otto and to be dependent only on the compression ratio. How¬ 
ever, as a steady-flow device, the compression ratio of the axial-flow 
compressor is better expressed in terms of the limiting pressures, rather 
than the limiting volumes, and, since Ti/T 2 = (pi/p 2 ) (fc-1)/fc , we may 
write 




( 12 : 12 ) 


The possibilities of the Brayton cycle, translated into the practical form 













288 


BASIC ENGINEERING THERMODYNAMICS 


of the gas-turbine power plant, seem to be tremendous. Not only does 
the air-standard cycle equal in efficiency the efficiency of the air-standard 
Otto, but the gas-turbine power plant is not limited to low compression 
ratios, since the compressor handles air alone and not a combustible 
mixture; nor is volatility a necessary property of the fuel, and a selection 
may be made from a wider range of fuels. What is perhaps an even more 
important advantage is based on the limited power which the reciprocat¬ 
ing internal-combustion engine may develop. Because the intermittent 
character of the flow of the working fluid through them limits the mass of 
working fluid which they can accept, Otto, diesel, and compression- 
ignition engines are bulky and expensive and are limited to comparatively 
low capacities. -The gas-turbine power plant, on the other hand, does 
not suffer from these limitations. 

The advantages suggested in the foregoing paragraph have not, at the 
present time, been fully realized because of practical factors, the most 
important of which is the metallurgical limit. Progress is being made in 
the improvement of the gas-turbine power plant along lines such as will 
be suggested later in our discussion; many of these improvements are 
based directly on thermodynamic principles. The reciprocating internal- 
combustion power plant may utilize, by proper design, the maximum 
temperatures within the cylinder which are available to the engineer as a 
result of the combustion process. This is possible partly because the 
piston and internal surfaces of the cylinder are exposed to these tem¬ 
peratures over only a small fraction of the entire cycle; during the balance 
of the cycle, temperatures are lower, and these parts have an opportunity 
to pass on the heat which they receive. But the temperature at entry to 
the turbine of the gas-turbine power plant is steadily maintained; more¬ 
over, the cooling of the turbine blades is much less effective than the 
cooling of an engine piston, and they attain practically the same tempera¬ 
ture as the gases which flow over them. At present, metallurgical 
research has developed blades which can, as an extreme limit, retain 
necessary strength and resistance to corrosion at temperatures of around 
2000°F, and the temperature at entry to the turbine must be held below 
this level. It will be noted that this temperature is far above the accepted 
metallurgical limit for the usual materials with which the engineer works. 

If the compressor and turbine of the gas-turbine power plant could be 
built to carry out reversible (isentropic) processes, the limitation in the 
maximum temperature of the cycle would, as is indicated by Eq. (12:12), 
have no effect on the efficiency of the cycle. The two Brayton cycles 
shown in Fig. 12:12, a Mollier diagram, which differ only in the amount 
of heat supply and, consequently, the temperature of the gas as it flows 
through the turbine, would thus have the same efficiency. Instead, 
best present-day practice limits the maximum attainable efficiency of 


POWER-GAS SYSTEMS 


289 


either compressor or turbine to a maximum of about 85 per cent, and 
there seems to be little chance for any considerable increase above that 
level. Indeed, it is only because of comparatively recent improvements 
in compressor and turbine performance that the gas-turbine power plant 
has been able to enter into serious competition with the internal-combus¬ 
tion reciprocating engine. The irreversibilities encountered in flow 
through the compressor and turbine may be shown on the Mollier diagram 
as a shifting of points 2 and 4 to 
higher entropies, as was* brought 
out in Chap. 11 and as is illustrated 
in Fig. 12:12 by the position of 
points 2' and 4'. Thus the net 
work of the cycle is (h s — hv) 

— (hzt — hi), a smaller amount 
than for the reversible cycle, both 
because the turbine work is less and 
because the work of compression is 
greater than for that cycle. It will 
be readily appreciated that, if the 
irreversibilities are serious, it is 
possible that no net work may be 
realized. Moreover, this effect is 
relatively more serious, at given 
compressor and turbine efficiency, 
for the smaller cycle of Fig. 12:12 than for the larger, because the 
elevation of point 2' above point 2 is based not upon the area of the 
Brayton cycle but upon the work of the compressor, and, similarly, 
the amount by which h# is greater than hi depends upon the gross work 
of the turbine. If higher temperatures could be maintained in flow 
through the turbine, corresponding to the larger cycle of the figure, the 
loss in the net work of the cycle, in other words, would be relatively less 
serious and the efficiency of the real gas-turbine power plant would be 
greater. It would appear that one of the most profitable fields for 
research toward the improvement of the efficiency of the gas-turbine 
power plant lies in the development of alloys still more heat-resistant 
than those at present available. 

Example 12:8A. A Brayton air-standard cycle operates between pressures of 
14 and 140 psia, and the lowest and highest temperatures of the cycle are 90 and 
3000°F, respectively. Calculate (a) the temperatures at the end of compression 
and at the end of expansion, ( b ) the heat supplied, heat rejected, and net work per 
pound of air, and ( c) the efficiency, (d) If the compressor and turbine processes are 
each assumed to be carried out at 85 per cent efficiency, what is the effect on the 
efficiency of the cycle? 



Fig. 12:12. Gas-turbine power-plant 
cycle—effect of metallurgical limit. 







290 


BASIC ENGINEERING THERMODYNAMICS 


Solution: 


(а) From data: px = p\ = 14 psia; p 2 = pz = 140 psia; T i = 550°R; T 3 = 3460°R. 

T 2 = Tx y k 1)/k = (550)(10)°- 4/1 - 4 = (550)(1.93) = 1060°R 

t ‘- t 'Y 2 = 550 (iolo) = 1790 ° R 

(б) Qs = c p (T* - T 2 ) = 0.24(3460 - 1060) = 576 Btu/lb 

Qr = CpiTi - Tx) = 0.24(1790 - 550) = 298 Btu/lb 

Net work = Qs — Qr 


Checking, 

Qs — Qr 


(k-l)/k 


= 278 Btu/lb 

/ 1 \ 0 . 4 / 1.4 

= 1 _ = 1 - 0.518 = 0.482 


v = 


Qs 


278 

576 


= 0.482 


(d) Based on a compressor efficiency of 0.85 and using the notation of Fig. 12:12, 

550 - 1060 


h '~ hl = 0.85 or Cp{T ' ~ Tt) 


hx — h 2 
For the turbine, 

/13 — hi 


= 0.85 or 


c P {Tx - Tv) 550 - Tv 


c p (T 3 - Tv) 3460 - Tv 


= 0.85 and Tv = H50°R 


= 0.85 and Tv = 2040°R 


h z - hi ~ c p {T 3 - T 4 ) 3460 - 1790 

Compressor work = c v (Tx — Tv) = 0.24(550 — 1150) = —144 Btu/lb 
Turbine work = c p (Tz — Tv) = 0.24(3460 - 2040) = 340 Btu/lb 

Net work = 196 Btu/lb 

Qs = c p {Tz - Tv) = 0.24(3460 - 1150) = 554 Btu/lb 


net work 196 


V = 


Qs 


554 


= 0.354 


The cycle efficiency has been reduced by slightly more than one-fourth by introducing 
the irreversibilities of the compressor and the turbine into the problem. 

Example 12:8 B. Solve the problem of Example 12:8 A, obtaining the same quan¬ 
tities, but based on a maximum temperature of 2000°F (instead of 3000°F). 

Solution: 

(а) From data: px = Pi = 14 psia; Tx = 550°R; Tz = 2460°R; T 2 = (550)(10)°- 4/l - 4 
= 1060°R; Ti = Tx (Jfy = 550(Hfff) = 1275°R. 

(б) Qs = 0.24(2460 - 1060) = 336 Btu/lb 
Q r = 0.24(1275 - 550) = 174 Btu/lb 

Net work = Qs — Qr = 162 Btu/lb 


(c) 77 = 

(d) 


Qs - Qr 162 


Qs 

Tx - T 2 


Tx - Tv 
Tz - Tv 
Tz - Ti 


336 
550 - 1060 
550 - Tv 
2460 - Tv 


2460 - 1275 


= 0.482 

= 0.85 and Tv = 1150°R 
= 0.85 and Tv = 1453°R 





















POWER—GAS SYSTEMS 


291 


Compressor work = 0.24(550 - 1150) = -144 Btu/lb 
Turbine work = 0.24(2460 - 1453) = 242 Btu/lb 

Net work = ~98 Btu/lb 

Qs = 0.24(2460 - 1150) = 315 Btu/lb 


net work _ 98 
~Qs “ 315 


0.312 


Note that the effect on cycle efficiency of including compressor and turbine irrever¬ 
sibilities has been more serious here than in Example 12:8 A. 


12:9. Improvements in the Gas-turbine Power-plant Cycle. While 
awaiting the products of continuing metallurgical research, there are 
some steps that can, and are, being taken with a view toward increasing 
the efficiency of the gas-turbine power plant. Essentially, these consist 
of various applications of the principle of regeneration. The gas-turbine 
power plant, as a steady-flow device, is well adapted for the employment 



Fig. 12:13. Regeneration in the gas-turbine power plant. 


of regeneration without the necessity of including complicated heat 
exchangers of the type described in Art. 4:6. The temperature at point 
4 of the Brayton cycle is often above that at point 2. In the gas-turbine 
power plant, these points represent, respectively, the states of the fluid 
as exhausted from the turbine and as discharged from the compressor; 
the thought occurs that some of the fuel can be saved by utilizing the 
stored energy in the exhaust gases to preheat the air before it enters the 
combustion chamber. Note that this preheating should be done after, 
rather than before, compression, or the work of compression will be 
increased. The flow arrangement is diagramed in Fig. 12:13. In the 
limit never reached in practice, the use of counterflow heat exhange, as 
shown in the figure, would permit the compressed air to be raised in tem¬ 
perature to Ta, while the turbine exhaust cools to T 2. The effect upon 
the air-standard cycle is shown in Fig. 12:14, where the crosshatched 














292 


BASIC ENGINEERING THERMODYNAMICS 


areas are equal and represent the heat exchanged. The net work of the 
air-standard cycle is not affected, but the heat supplied from the source 

is reduced in the proportion 
(T 3 - T a )/{Ts - T 2 ), with a con¬ 
sequent improvement in cycle ef¬ 
ficiency. 

Example 12:9 A. Apply regeneration 
to the cycle of Example 12:85, part d, 
and show the effect on the efficiency of 
the cycle. Assume regeneration to be 
ideal, i.e., the air enters the combustion 
chamber at the temperature TV. 

Solution. This would have no effect on 
the compressor or turbine work. The 
heat supplied would be reduced to 

Qs = c p (Tz - TV) = 0.24(2460 

- 1453) = 242 Btu/lb 
net work 98 

V = 



Fig. 12:14. Bray ton cycle 
regeneration. 


effect of 


Qs 


242 


= 0.405 


Other steps that may be taken to¬ 
ward the attainment of higher effi¬ 
ciencies of the gas-turbine power plant include the compression of the air 
in two (or more) compressors in series, with intercooling of the air between 


Air 



exchanger 

Fig. 12:15. The gas-turbine power plant—staged compression and expansion. 


compression steps, and the similar division of the expansion into two 
steps, the gases being reheated after leaving the high-pressure turbine by 
passing them through a secondary-combustion chamber to raise their 
temperature before continuing their expansion in the low-pressure turbine 
unit. Staged compression with intercooling, or staged expansion with 
reheat, may be applied to the gas-turbine power-plant cycle separately 


































POWER—GAS SYSTEMS 


293 


or in combination with each other but, for reasons that will later develop, 
should always be used in combination with regeneration, as described in 
the earlier part of this article. The flow arrangement, when both are 
introduced into the cycle, is diagramed in Fig. 12:15; note that the 
regenerative heat exchange takes place between the gases as they are 
exhausted from the low-pressure turbine and the air as it leaves the final 
stage of compression. 

Exclusive of the effect of regeneration, the changes introduced into the 
air-standard cycle are illustrated in Fig. 12:16. In this figure, the basic 
Brayton cycle is shown as l-c-3-/. T 
The effect of staged compression with 
intercooling is indicated by the line 
l-a, representing the first stage of 
compression, the line a-b, showing the 
effect of intercooling, and the line b- 2, 
the second stage of compression. As 
a result of staging the compression, 
the work area 2 -c-a-b is saved and 
therefore, in effect, added to the net 
work of the cycle, which is now traced 
as l-a-6-2-3-/. If the cycle is revers¬ 
ibly performed, without regeneration, 1 

it will be noted that this extra work is F 

e 

not a free dividend but must be paid 
for in terms of the extra heat, represented by area 2 -c-h-g, that must be 
supplied the working fluid per cycle. With regeneration, because of the 
lower temperature at point 2, this extra heat may be recovered from the 
exhaust of the turbine and the heat supplied by the source (from/tteZ, in the 
real plant) need not increase; a net gain in efficiency is the result. 

Turning now to the effect of staged expansion, with reheat, it is 
observed that a secondary combustion can be carried out in the real plant 
because it has been necessary to keep the temperature of the gases, as they 
enter the turbine, below the metallurgical limit and thus well below the 
maximum temperature that could result from combustion; this has been 
accomplished by using, in the first stage of combustion, a large amount of 
air in excess of that necessary to provide oxygen for the combustion of the 
fuel. The exhaust from the high-pressure turbine thus contains con¬ 
siderable uncombined oxygen and can support further combustion. 
Referring again to Fig. 12:16, it is noted that an extra dividend of work is 
secured by staged expansion, with reheat, which is represented by the 
area d-e-4-f; this comes at the expense of the extra heat supplied from the 
source (the fuel), as represented by area d-e-j-i. Without regeneration, 
the efficiency of the cycle would be lowered, for the ratio of these areas is 



r iG. 12:16. The gas-turbine power- 
ilant cycle—staged compression and 
xpansion. 











294 


BASIC ENGINEERING THERMODYNAMICS 


less than the efficiency of the basic Brayton cycle. But, with regenera¬ 
tion, the temperature of the exhaust gases is raised enough so that the 
exchange of heat due to regeneration is increased to the point where most 
of this extra heat need not come from the fuel but is saved by causing the 
air to enter the first stage of combustion at a higher temperature. In 
other words, a large part of the fuel used in the secondary combustion is 
saved in the primary combustion process. 


Example 12-9 B. Staged compression, with intercooling, and staged expansion, 
with reheat, are used in a gas-turbine power plant that operates between the over-all 
pressure limits 14 to 140 psia and between extreme temperatures of 90 and 2000°F, 
as in Example 12:82?. Two stages of compression and two stages of expansion are 
employed, as in Fig. 12:16, and the intermediate pressure is such that the work of 
compression and the work of expansion are divided equally between stages. Inter¬ 
cooling reduces the temperature of the air to 90°F at entrance to the second stage of 
compression, and reheat between turbine stages is to 2000°F. Compressor and tur¬ 
bine efficiencies are 85 per cent for each stage. Calculate the effect on cycle efficiency 
(a) without regeneration and (6) with ideal regeneration. 

Solution. For equal division of work between stages in both compression and 
expansion, pi/pi = (inr)* = 3.16 and pi = (3.16) (14) = 44.3 psia. With reference 
to Fig. 12:16, 7\ = 550°R; T a = (550)(3.16) 0 - 286 = (550)(1.39) = 765°R; T b = 550°R; 
T 2 = (550)(3.16) 0 - 286 = 765°R; T d = 2460(1/3.16) 0 - 286 = 1770°R; T e = 2460°R; T< 
= 2460(1/3.16) 0 - 286 = 1770°R. 


(Ideal) compressor work is (2) (0.24) (550 — 765) = —103 Btu/lb 
(Ideal) turbine work is (2)(0.24)(2460 — 1770) = 331 Btu/lb 
Actual compressor work is —103/0.85 = —121 Btu/lb 
Actual turbine work is (0.85)(331) = 281 Btu/lb 

Net work = 160 Btu/lb 


The actual temperatures at exit from the first and second stages of the turbine are 
higher than T d and T 4 as calculated above. They may be computed from the relation 


T 3 - 7V 
T 3 - T d 
T e — TV 

Te - T< 


2460 - T d ’ 
2460 - 1770 
2460 - TV 
2460 - 1770 


= 0.85 or T d - = 1873°R 
= 0.85 or Tv = 1873°R 


Similarly, the temperature of the air at exit from the second stage of compression is 
designated as 7V, where 


T b - T 2 
T b - T v 


550 - 765 
550 - T r 


0.85 or T v = 803°R 


TV 


(a) Without regeneration, 


Heat supplied in primary combustion = c p (T 3 — 7V) = 0.24(2460 — 803) 

= 396 Btu/lb 

Heat supplied in secondary combustion = c v (T e — TV) 

= 0.24(2460 - 1873) = 141 Btu/lb 

Qs = 537 Btu/lb 


net work 160 


Qs 


537 


= 0.298 


v 












POWER—GAS SYSTEMS 


295 


Note that this is a lower cycle efficiency than was calculated in Ex. 12:85. 

(6) With ideal regeneration, the energy stored in the exhaust gas can be used to raise 
the temperature of the air as it enters primary combustion to 7V(= 1873°R). Then 


Heat supplied in primary combustion = c p (T* — TV) = 0.24(2460 — 1873) 

= 141 Btu/lb 

Heat supplied in secondary combustion = c p (T e — T <*') 

= 0.24(2460 - 1873) = 141 Btu/lb 

Qs = 282 Btu/lb 


v = 


net work 160 


Qs 


282 


= 0.567 


This result, when compared with the efficiencies obtained in Example 12:8 B and 
Example 12:9A, indicates that staged compression and expansion, when used in com¬ 
bination with regeneration, may have a beneficial effect on the efficiency of the cycle. 


12:10. Jet Propulsion. The gas-turbine power plant has assumed 
increasing importance in the field of airplane propulsion. What had 
come to be regarded as the conventional method of propulsion consisted 
of an airscrew, or propeller, to which power was furnished by an internal- 
combustion engine of the reciprocating type. The efficiency of the pro¬ 
peller dropped off sharply as the speed of its tips approached the velocity 
of sound, about 750 mph in air at sea-level temperature, less at the higher 
altitudes. Because of the composition of the rotative velocities of the 
propeller segments with the forward (translational) velocity of the air¬ 
plane, these tip speeds exceeded considerably the translational velocity, 
and the propeller-driven aircraft was limited in its efficient operation to 
speeds below 500 mph. By utilizing the gas-turbine power plant, it was 
possible to by-pass this “roadblock.” In this application, the expansion 
in the turbine is halted at a point where the turbine can deliver just suffi¬ 
cient power to drive the compressor; the gases, as they leave the turbine, 
are at a pressure and temperature which is elevated above that corre¬ 
sponding to point 4 of Fig. 12:9. Because of their elevation of pressure 
and temperature, these gases may be ejected rearward through a nozzle¬ 
shaped opening at high velocity, and their consequent change of momen¬ 
tum results in a forward-acting reaction force (thrust) on the airplane. 
The propulsive efficiencies attained by this method of providing thrust 
increase with airplane speed, and the handicap of the propeller is thus 
avoided. 

The airplane gas-turbine power plant may, of course, be adapted for 
propeller drive by continuing the expansion to atmospheric pressure. A 
variant is found in the possibility of changing the limit of expansion so 
that power may be delivered to the propeller at low airplane speeds, when 
propeller drive is more efficient, and providing that, at the higher speeds, 
jet propulsion may take over. When the expansion in the turbine is 
intermediate between these two extremes, the propeller and the jet may 






296 


BASIC ENGINEERING THERMODYNAMICS 


operate concurrently to provide the thrust that pulls the airplane along 
its flight path. 

At least in the present stage of its development, the gas-turbine power 
plant is considerably less efficient than the reciprocating-engine type of 
airplane power plant, and this means that a greater weight of fuel must be 
carried to accomplish the same task; this is a handicap of considerable 
importance in airplane design, where weight has great significance. How¬ 
ever, a part of this extra weight is saved in the weight of the power plant 
itself; the gas-turbine power plant weighs much less per unit of rated 
power than does the reciprocating engine. 

The gas-turbine power plant, as a jet-propulsion device, may be classi¬ 
fied as of the continuous-firing air-stream type. An example of the 
intermittent-firing air-stream jet-propulsion engine is the V-l bomb (the 
“buzz bomb’’) used by the Germans in the Second World War. This is 
shaped in the form of a nozzle and is given a considerable initial velocity 
by being projected into the air by a launching mechanism with the axis of 
the nozzle along the line of flight. Fuel is injected into the air-stream 
ahead of the throat of the nozzle, and the resulting mixture is exploded 
by means of an electric spark. The backflow of the exploded charge is 
prevented by shutters, which are located at the forward end of the duct 
and which close at once when backflow starts. The exploded charge then 
expands to the rear through the throat and, owing to its increased momen¬ 
tum, provides the forward thrust that is necessary to maintain the speed 
of the bomb. When the exploded charge has been dissipated, the ram 
action of the air stream again opens the shutters and the cycle is repeated. 

When the nozzle is projected through the air at velocities in excess of 
the velocity of sound, the shutters of the buzz bomb are no longer neces¬ 
sary; if the flow at the throat of the nozzle is supersonic, it has been shown 
in Chap. 11 that backflow cannot take place. This type of device is 
called an athodyd (from aero-thermodyneanic-duct ); like the gas-turbine 
jet engine, the athodyd is a continuous-firing device. 

In both buzz bomb and athodyd the supply of heat from the source (the 
fuel) begins at a temperature not far from that of the atmosphere. The 
average temperature at which heat is supplied is thus lower, and the 
attainment of high efficiency in the utilization of fuel is handicapped in 
comparison with the gas turbine. 

The rocket is a jet-propulsion device which carries not only fuel but also 
oxygen for its combustion. These reactants combine progressively, and 
the products of combustion are ejected rearward at high velocity. The 
weight of oxygen carried is a large part of the total weight of the rocket, 
and the performance of this type of jet-propulsion device is correspond¬ 
ingly handicapped in comparison with the air-stream type described 
above; however, it is not dependent on oxygen from the surroundings and 


POWER—GAS SYSTEMS 


297 


thus may operate outside the earth’s atmosphere. The charge of react¬ 
ants in the head of the rocket may be in the form of solids, as in the 
Fourth-of-July skyrocket, or may be liquids, as in the V-2 bomb. In 
either case, provision must be made for progressive, rather than instan¬ 
taneous, combustion of the charge. 

12:11. The Holzwarth Cycle and the Explosion Gas Turbine. The 
same reasoning that was used in extending the expansion process of the 
diesel cycle to form the Brayton cycle may be applied to the Otto cycle. 
The result is known as the Holzwarth cycle and is illustrated in Fig. 12:17. 
1 his cycle differs from the Brayton cycle in the substitution of constant- 
volume, in place of constant-pressure, combustion. Some attempts have 




Fig. 12:17. Holzwarth air-standard cycle. 

been made to utilize this cycle as the basis of a real power plant, and these 
have, in general, taken the form of an explosion gas turbine. This device 
includes a combustion chamber into which a charge of air is forced through 
the intake valve. Fuel is sprayed into this air charge through a fuel noz¬ 
zle, and, with all valves closed, a spark ignites the mixture, exploding it. 
The exhaust valve then opens and permits the hot products of combustion 
to flow through nozzles and act on the blades of a turbine wheel, until the 
flow stops owing to a decrease of pressure. The action is intermittent, 
the gases impinging on the turbine buckets at variable velocity. This is 
not, as we have seen in Chap. 11, a condition that conduces to high turbine 
efficiency. 

12:12. External Combustion and the Closed Cycle. Internal combus¬ 
tion is, of course, out of the question if the working fluid of the real engine 
is to follow the closed, controlled cycle of the heat engine. The classic 
example of the closed-cycle heat engine, using a gas as the working fluid, 
is the Carnot. The Carnot engine is purely an idealization for reasons 











298 


BASIC ENGINEERING THERMODYNAMICS 


that have been fully covered in preceding pages. Three of the principal 
reasons will bear restatement at this time: 

1. Heat is supplied to and rejected from the Carnot engine at constant 
temperature. This is not a practical method of heat exchange for a gas. 

2. The net work of the Carnot engine is extremely small in comparison 
with the work of expansion and of compression; in other words, the mean 
effective pressure of the cycle is very low. 

3. The working cylinder of the Carnot engine must perform both the 
function of a prime mover and of a heat exchanger, an impractical com¬ 
bination of duties in the real engine. 




Fig. 12:18. Idealized cycles of the air engines. 

These reasons are entirely sufficient to rule out the use of a real Carnot 
engine which uses a gas as the working fluid. They do not alter the value 
of the Carnot engine as a speculative device, an imaginary type of heat 
engine with which the ultimate in efficiency is associated. Let us, how¬ 
ever, pass on to the cycles of real engines which can pass the gas system 
through a closed chain of processes, these engines thus being complete 
heat engines. 

The Regenerative Cycles. The Stirling and the Ericsson cycles are 
shown in Fig. 12:18. These are the idealized forms of the cycles that are 
the basis of operation for the Stirling and the Ericsson air engines. Both 
cycles receive heat from the source only during the isothermal process 1-2 
and reject heat to the refrigerator only during the isothermal 3-4. Each 
cycle rejects heat temporarily during the processes which are designated 
by the paths 2-3 of Fig. 12:18, but this heat is rejected to a regenerator 
which replaces it in the working fluid during process 4-1. Constant- 
volume regeneration, as described in Art. 4:6, is used in the Stirling cycle, 
regeneration at constant pressure in the Ericsson cycle. Both cycles may 
be shown to have an efficiency equal to ( T s — T R )/T S and thus to the 








POWER—GAS SYSTEMS 


299 


limiting standard set by the Carnot cycle; this has already been demon¬ 
strated for the Stirling cycle (see Chap. 4). 

The real Stirling and the real Ericsson engines both include a working 
and a displacer cylinder and piston. To illustrate why these engines fall 
far short of the performance predicted by a thermodynamic analysis of 
their idealized cycles, the scheme of operation of the Stirling engine will 
be described below. 

The displacer cylinder of the Stirling engine is closed at top and bottom, 
the source of heat (usually an open flame) being in contact with the bot¬ 
tom head and the refrigerator (either the open atmosphere or cooling 
coils) with the upper head. The displacer piston, as it moves vertically 
through its stroke, neither performs nor receives work but merely dis¬ 
places the air from the bottom of the cylinder to the top, or vice versa, 
through side passages. These passages contain metal surfaces which 
exchange heat with the air and thus are capable of acting as regenerators. 
A passage leads from the displacer cylinder into the adjacent working 
cylinder, entering that cylinder beneath the working piston, and between 
that piston and the closed end of the working cylinder. At point 1 
(Fig. 12:18) of the Stirling cycle, the displacer piston is at the upper end 
of its stroke and the working piston at the bottom of its cylinder. The 
volume of the enclosed air system is that of the displacer cylinder and 
regenerator passages, less the volume of the displacer piston. The air is 
at temperature T s and is in contact with the lower head of the displacer 
cylinder; it therefore receives heat from the source through that head. 
This heat causes an expansion of the air, and the excess volume passes to 
the working cylinder and does work on the working piston as it moves 
through its upward stroke; by proper adjustment of the rate of movement 
of the working piston to the rate of heat supply, this expansion may be 
conceived as taking place at constant temperature. When the working 
piston reaches the top of its stroke, process 1-2 has been completed and 
the system has a volume V 2 , greater by the displacement of the working 
piston than V h but still is at the temperature T s ; thus p 2 < Pi. When 
point 2 is reached, the displacer piston, which has remained stationary at 
the top of its cylinder during the power stroke 1-2 of the working piston, 
begins to move downward; the working piston is now stationary. As it 
falls, the movement of the displacer piston produces no change in the 
total volume of the air system; the air is merely displaced from the bottom 
to the top of that piston, passing through the regenerator in transit. The 
material of the regenerator is cool and absorbs heat from the air so that 
this air issues into the space atop the displacer piston at the lower tem¬ 
perature T r . When the downward stroke of the displacer piston has been 
completed, the air of the system has the volume V z (equal to V 2 ) and, 
because the temperature has been decreased, the lower pressure w 



300 


BASIC ENGINEERING THERMODYNAMICS 


The air is now in contact, through the upper head of the displacer cylin¬ 
der, with the refrigerator, and heat begins to flow from the enclosed sys¬ 
tem to the refrigerator. This is the beginning of process 3-4, which con¬ 
tinues with the displacer piston stationary at the bottom of its cylinder 
and the working piston falling at a rate such that the temperature of the 
air remains constant at T R . When the working piston has reached the 
end of its return stroke, the volume is again the same as at point 1, but, 
because the temperature is T R , the pressure p 4 < Pi- The only step 
necessary to return to point 1 at which we started the cycle is for the dis¬ 
placer piston to return to the top of its cylinder, with the working piston 
stationary at the bottom of its travel. This returns the air through the 
regenerator, which is hot because of the heat it received during process 
2-3, and the air system issues into the space below the displacer piston 
(and in contact with the source through the lower head of the displacer 
cylinder) at the temperature T s . Note that, in causing this temperature 
rise in the air, the regenerator has been cooled and is in a condition to per¬ 
form its function of cooling the air during process 2-3, as the cycle is 
retraversed. 

In the foregoing description, the operation of the Stirling engine has 
been highly idealized. The real engine obtains an approximation of 
alternate movement of displacer and working pistons by connecting them 
to the engine shaft through a 90° bell-crank arrangement, the displacer- 
piston crank being 90° ahead of the working-piston crank; this means that 
each piston, in turn, is not absolutely stationary while the other carries 
out its stroke. We have already observed that the isothermal process, 
not being subject to close control, is not a practical method of transferring 
heat to or from a gas system. There are large inherent irreversibilities 
connected with the practical application of the process of regeneration. 
The use of the displacer cylinder as a heat exchanger limits the perform¬ 
ance of the engine by limiting the rate of heat transfer, and so the capacity 
of the air engine; engines of this type are exceedingly large in comparison 
with the power they may develop, and this factor has the effect of increas¬ 
ing the relative magnitude of losses due to mechanical friction. The 
effect of these considerations is to cause the efficiency of the real air engine 
to be only a small fraction of that predicted from an analysis of its idealized 
cycle. Air engines are not of present-day importance as prime movers, 
but they do exemplify a type of real engine, using a gas as the working 
fluid, that is a heat engine in principle. Being simple, and largely 
foolproof, in operation, they still are used in small sizes. 

The Closed-cycle Gas-turbine Power Plant. External combustion may 
be applied to the gas-turbine power plant, as shown in the flow diagram of 
Fig. 12:19, and, combined with a cooling of the exhaust gases to atmos¬ 
pheric temperature, as also shown in that figure, can close the cycle and 


POWER—GAS SYSTEMS 


301 


convert the plant to a complete heat engine. If the heat-rejection process 
is omitted, the cycle is said to be semiclosed. 

Ihe advantages which can be seen in the application of the closed cycle 
to the gas-turbine power plant include: 

1. The use of a cheaper fuel, such as coal, is possible. 

2. A clean working fluid passes through both compressor and turbine, 
thus reducing damage to blading due to erosion and corrosion. 

3. Higher pressures may be used at entrance to the compressor and 
around the cycle, making possible a reduction in plant size for a given 
power. 



Fig. 12:19. The closed-system gas-turbine power-plant cycle. 


4. An inert gas may be substituted for air, further lessening the likeli¬ 
hood of corrosion. 

5. Fractional loads may be handled at high efficiency by varying the 
density of the gas system; this can be accomplished by providing for addi¬ 
tion or subtraction of gas as the load demands. 

The disadvantages may be listed as: 

1. Additional complexity and consequent higher cost are incurred. 

2. The introduction of a heat exchanger to replace the more efficient 
internal-combustion process is necessary. 

3. A coolant is required. 

4. The metallurgical limit imposes a lower maximum cycle tempera¬ 
ture, because of the substitution of a heat exchanger for the combustion 
chamber. 


























302 


BASIC ENGINEERING THERMODYNAMICS 


Problems 

1. A test of an airplane engine which operates on the Otto cycle with a compression 
ratio of 8 shows it to be capable of delivering a horsepower-hour for an expenditure of 
0.5 lb of gasoline (heat value = 18,500 Btu/lb). What percentage of air-standard 
efficiency is realized? 

2. Work Example 12:3, changing the limiting temperatures of the cycle to 70 and 
2500°F, respectively, but leaving other data unchanged. Compare your answers with 
those of the example, and comment on the effect of the change in data. 

3. At the beginning of compression in an air-standard Otto cycle, Vi = 1 ft 3 , 
t x = 100°F, pi = 15 psia. The temperature at the end of compression is 650°F, and 
the pressure p 3 is 300 psia. Calculate the heat supplied, heat rejected, net work, 
air-standard efficiency, mean effective pressure, clearance, and compression ratio for 
this cycle. 

4. The efficiency of an air-standard Otto cycle is 54 per cent. What is its clearance? 

5. Plot the efficiency of the air-standard Otto cycle vs. its compression ratio for 
ratios between 4 and 12. 

6. The pressure at the beginning of the compression stroke in an air-standard Otto 
cycle is 14 psia. At the end of the stroke it is 130 psia. What is the efficiency of 
the cycle? 

7. Using the Ts diagram as a tool, show the effect of compression ratio on the 
efficiency of the air-standard Otto cycle. 

8. To what cutoff percentage is a load ratio of 2 equivalent in a diesel engine with 
a compression ratio of 15? A compression ratio of 12? A compression ratio of 18? 

9. At the beginning of compression in an air-standard diesel cycle, pi = 15 psia, 
ti = 100°F, and V x = 1 ft 3 . The pressure at the end of compression is 600 psia. 
The temperature f 3 = 3000°F. Calculate the heat supplied, the heat rejected, net 
work, air-standard efficiency, mean effective pressure, clearance per cent, cutoff 
per cent, and load ratio for this cycle. 

10. Calculate the air-standard efficiency of a diesel cycle having a load ratio of 2 
and a compression ratio of 14. 

11. Calculate the air-standard efficiency of a diesel cycle having a cutoff percentage 
of 5 and a compression ratio of 15. 

12. Based upon a load ratio of 2, plot the efficiency of the air-standard diesel cycle 
vs. its compression ratio for ratios between 8 and 18. Compare with the similar plot 
for the Otto air-standard cycle (see Prob. 5), and state your conclusions. 

13. A diesel cycle has an air-standard efficiency of 0.60 when the load ratio is 1.8. 
What is its compression ratio? 

14. Derive an expression for the efficiency of the air-standard dual cycle in terms 
of the absolute temperatures at points 1, 2, 3, 4, and 5. Based upon this expression, 
discuss the factors that affect the air-standard efficiency of this cycle. 

15. For an air-standard dual cycle, p x = 14 psia, t x = 90°F, V x = 2 ft 3 , V 2 = 
0.15 ft 3 . Twenty Btu of heat is received by the charge at constant volume and 30 Btu 
at constant pressure. What are the pressure, temperature, and volume of the charge 
at the end of the adiabatic expansion? Calculate the heat rejected, net work, air- 
standard efficiency, compression ratio, clearance per cent, cutoff per cent, and mean 
effective pressure for this cycle. 

16. Compare the Otto, diesel, and dual air-standard cycles on Ts coordinates on 
the basis of equal compression ratios. State your conclusions. Would you consider 
this a fair and reasonable basis of comparison? 

17. Derive an expression for the efficiency of an air-standard Lenoir cycle in terms of 
the absolute temperatures at points 2, 3, and 4. Apply this expression to the calcula- 


POWER—GAS SYSTEMS 


303 


tion of the efficiency of a Lenoir air-standard cycle for which fa = 90°F and fa = 
3000°F. Compare with the efficiencies calculated in Examples 12:3, 12:4, and 12:5 
for Otto, diesel, and dual air-standard cycles operating between the same temperature 
extremes, and state your conclusions. Use the Ts diagram to make the same 
comparison. 

18. Work Example 12:8A with the upper pressure of the cycle changed to 200 psia, 
all other data the same. Compare with the results obtained in the solution of the 
example, and discuss. 

19. Assuming turbine and compressor efficiencies to be equal, how low could they 
fall under the conditions of Example 12:8A before the net output of the plant would 
vanish? 

20. Work Example 12:85 with the upper pressure of the cycle changed to 200 psia, 
all other data the same. Compare with the results of the example and discuss. 

21. Assuming turbine and compressor efficiencies to be equal, how low could they 
be under the conditions of Example 12:85 before the net ouput of the plant would 
become zero? 

22. In Example 12:8A, how low could the highest temperature of the cycle drop 
before the net output of the plant would vanish when the compressor and turbine 
efficiencies were both 0.85? 

23. Work Example 12:85, part d, but change the compressor and turbine efficiencies 
to 0.80. Compare with the answer obtained in the solution of the example, and discuss. 

24. Apply ideal regeneration to the cycle of Prob. 20, part d. Compare the cycle 
efficiencies with and without regeneration and discuss. 

25. Apply ideal regeneration to the cycle of Example 12:85, part d, but with com¬ 
pressor and turbine efficiencies of 0.80 (see Prob. 23). Calculate the efficiency of the 
cycle, and compare with the answers obtained in Example 12:9A and Prob. 23, dis¬ 
cussing the effect of the change. 

26. Work Example 12:95, but change the upper pressure of the cycle to 200 psia. 

27. Develop an expression for the efficiency of the explosion-gas-turbine air- 
standard cycle in terms of the temperatures at points 1, 2, 3, and 4. Using the Ts 
diagram, compare this cycle with the Otto cycle of the same compression ratio with 
respect to their efficiencies. According to this diagram, what factor causes the Otto 
cycle efficiency to be exceeded? Is the efficiency of the explosion-turbine cycle solely a 
function of its compression ratio? Discuss the effect of changing the pressure and 
temperature rise during explosion on the air-standard efficiency of the cycle. 

28. For an explosion-turbine cycle, air-standard, pi = 14 psia, fa = 110°F. V x = 
2 ft 3 , p 3 = 300 psia, and the compression ratio V x /V 2 is 5. Calculate the heat supplied, 
heat rejected, net work, and efficiency, for this cycle. What is the efficiency of the 
equivalent air-standard Otto cycle? 

29. Show that the efficiency of the Ericsson cycle, with regeneration, will equal 
that of a Carnot cycle that operates between the same extremes of temperature. 


Symbols 

c a constant 

c p specific heat at constant pressure 
c v specific heat at constant volume 
h specific enthalpy 
J proportionality factor 
k ratio of the specific heats, c v /c v 
mep mean effective pressure 
M mass flow rate, per cycle or per unit time 


304 


BASIC ENGINEERING THERMODYNAMICS 


p pressure, psi 

P pressure, psf; pressure in general 
Q rate of heat flow, per cycle or per unit time 
R gas constant 
s specific entropy 
S entropy of a system 
T absolute temperature 
v specific volume 
V volume of a system 

W rate of work delivery, per cycle or per unit time 

Greek Letters 
77 efficiency 


Subscripts 

D displacement 
p constant pressure 
R refrigerator 
S source 

v constant volume 


CHAPTER 13 

THE RECIPROCATING STEAM ENGINE 


13:1. The reciprocating engine of the steam power plant is, like the 
turbine, not a heat engine since steam enters and leaves across its bounda¬ 
ries. It may take the place of the turbine in the plant and so form a part 
of the steady-flow heat engine described in Art. 3:8. It differs from the 
turbine, among other respects, in that the flow of steam is not steady and 
continuous, as in the turbine, but intermittent in character. However, 
the reciprocating engine follows a ma¬ 
chine cycle, and the over-all effect of its 
operation may be studied by applying 
the energ}^ equation of steady flow, as 
suggested in Art. 3:9. 

Some of the basic features of the recip¬ 
rocating steam engine are illustrated in 
Fig. 13:1. Steam is admitted to a cylin¬ 
der in which it operates to impart recip¬ 
rocating motion to a piston; this recip¬ 
rocating motion is converted into rotary 
motion through the action of a piston 
rod, crosshead, connecting rod, and 
crank. Referring to the illustration, the 
steam chest receives steam directly from 
the steam generator (the boiler, or boiler- 
superheater, unit of the steam-power- 
plant heat engine) and therefore contains 
steam at full boiler pressure while the engine is in operation. The 
inlet valves operate alternately to admit and to shut off the flow of 
live steam to the two sides of the piston; the exhaust valves alternately 
open and release the charge, so that it may flow into the exhaust line and 
thence to the condenser, and close, bringing to an end the escape of 
expanded steam from the cylinder. The cylinder, as is common in steam- 
engine design, is double-acting, i.e., as steam is admitted and expanded on 
one side of the piston, it is exhausted from the other end of the cylinder; 
thus a separate cycle of operation is carried out on each side of the piston, 
approximately doubling the capacity of a cylinder of given size. Double 
action is possible since the steam temperature is not high enough to cause 

305 

























306 


BASIC ENGINEERING THERMODYNAMICS 


cooling of the cylinder and piston to be a problem; deliberate cooling of 
cylinder surfaces, as in the internal-combustion engine, is not necessary, 
and the real steam engine is more nearly an adiabatic device than is the 
real internal-combustion engine. 

Considering individually the two cycles of operation that are carried 
out in the cylinder of a double-acting engine, the admission of live steam 
usually takes place through only a part of the power stroke; after the sup¬ 
ply has been cut off by the closing of the inlet, or steam, valve, the charge 
continues to expand until the piston reaches the end of its stroke. It is 
fully expanded only if the pressure has dropped to exhaust (condenser) 
pressure before the exhaust valve opens. Expansion in the real engine is 
seldom carried to this point since, as will be explained later, the losses 
outweigh the gains; thus, when the exhaust valve opens near the end of 
the power stroke, a rapid drop in pressure takes place as the partly 
expanded charge begins to escape into the exhaust passage. The exhaust 
valve remains open during the major portion of the return stroke of the 
piston; after it closes, the portion of the charge that remains in the cylin¬ 
der is compressed into the clearance space. As, or shortly before, the 
piston reaches its limit of motion on the return stroke (its dead-center 
position), the intake valve opens, a fresh charge of live steam begins to 
enter, and the cycle is retraced. The valves are opened and closed 
mechanically, being driven by a mechanism that operates off the engine 
shaft. The action of the valves is described in terms of the events which 
they control. The opening of the intake valve is called admission, and 
its location in the cycle is described in terms of the percentage of piston 
stroke yet to be covered before the dead-center position at the beginning 
of the power stroke is reached. Cutoff marks the closing of the inlet valve, 
its location being stated in terms of the percentage of the power stroke 
that has been traversed at the time the valve closes. Release is the name 
applied to the opening of the exhaust valve, and the position of the event 
is described on the same basis that is used for cutoff. Finally, the closing 
of the exhaust valve is termed compression , located as occurring at a per¬ 
centage of stroke determined in the same manner as for admission. Thus, 
in the location of all events of the cycle, the dead-center position of the 
piston at the beginning of its power stroke is conventionally used as the 
reference position. 

In the elementary form of the steam engine, the events are located at 
fixed and constant percentages of the stroke as long as the engine is in 
operation; in order to change their position, the engine must be shut down 
and the valve-operating mechanism adjusted. In more advanced design, 
the percentages at which the events occur may be controlled, either 
manually or automatically, while the engine is running. 

The type of engine cylinder illustrated in Fig. 13:1 is only one of the 


THE RECIPROCATING STEAM ENGINE 


307 


many forms in which the steam engine is designed and built. For exam¬ 
ple, here the inlet and exhaust valves are distinct and separate, each hav¬ 
ing a semirotary motion of its own, and the location of inlet and exhaust 
events may be controlled independently of each other; in a type which is 
more common in practice (called the slide-valve engine), a single valve 
controls all events on both ends of the cylinder, and the exhaust steam 
leaves through the same openings (called ports) as are used by the live 
steam as it enters. The inlet ports always lead to the ends of the cylinder, 
but when a separate exhaust valve is used, it may be placed at a consider¬ 
able distance from the cylinder head, as in the unijiow engine to be later 
described. 

13:2. The Reversible Adiabatic Engine. The indicator diagram of a 
reversible adiabatic engine, without clearance, is illustrated in Fig. 13:2. 
The intake valve opens at point 3, 
the pressure mounts to the supply 
pressure while the piston is still sta¬ 
tionary at the end of its stroke (3-4), 
and line 4-1 represents the induction 
of the charge as the piston begins its 
power stroke, point 1 being the point 
of cutoff. During this portion of 
the power stroke, the work performed 
on the piston is represented by the 
area 4-l-a-0; if the weight of the 
charge is designated as M , this area 
is MPiVi. After cutoff, and until 
release at point 2, the charge expands 
isentropically, doing work on the piston equal to the area 1-2 -b-cr, since 
this expansion is adiabatic, an application of Eq. (2:3) shows that this 
work area equals JM(u\ — u 2 ). Finally, the exhaust valve remains open 
during the return stroke, closing simultaneously with the opening of the 
intake valve (at point 3), and the work done by the piston (negative work) 
during this stroke is equal to the area 2-3-0 -b; this area is — MP 2 v 2 . 
Summing up the work areas, the net work of the cycle is MPiVi + JM(u\ 
— u 2 ) — MP 2 v 2 = JM(hi — h 2 ). Since points 1 and 2 lie on a line of 
constant entropy, we may designate the work of the reversible cycle as 

Frev cycle JM(hx - h 2 ) s 

This cycle, as a reversible cycle, delivers the maximum possible amount of 
work under the given conditions, and the ideal net work for the adiabatic- 
engine cycle is therefore, in Btu per pound of steam flowing through the 
engine, (hi — h 2 ) s , where the subscript 1 refers to the condition of the 
steam as it is supplied the engine and the subscript 2 to a state having the 



versible adiabatic engine without 
clearance. 







308 


BASIC ENGINEERING THERMODYNAMICS 


same entropy but a pressure equal to condenser (exhaust) pressure. It 
is noted that the reversible adiabatic engine fully expands the charge of 
steam; if the state of the steam as it is supplied the reversible engine and 
the exhaust pressure are both constant, the cutoff is fixed at a definite per¬ 
centage of the stroke of the reversible adiabatic engine. 

The introduction of clearance, necessary in the real engine to cushion 
the piston as it changes the direction of its motion, does not mean that a 



versible adiabatic engine, with clear¬ 
ance. 

reaches the end of its stroke. T1 
cent of piston displacement, or 


reversible adiabatic engine is no 
longer conceivable or that the ideal 
amount of work that may be per¬ 
formed per pound of steam supplied 
the engine will be any different. In 
Fig. 13:3 the cycle of a reversible 
adiabatic engine, with clearance, is 
shown by the full lines, the dashed 
lines corresponding to the equivalent 
cycle, without clearance, as in Fig. 
13:2. In this case the piston does not 
move into contact with the cylinder 
head but leaves the clearance volume 
Vb between it and the head as it 
amount of clearance is expressed in per 


Clearance, per cent 


lOOVb 

V D 


lOOVb 
V 2 - Vb 


( 13 : 1 ) 


In the reversible adiabatic cycle of Fig. 13:3, the exhaust valve closes at 
point 6, trapping a portion of the charge having the volume F 6 at the 
exhaust pressure. The weight of steam thus trapped is called the weight 
of cushion steam and will be designated as M c . As the piston continues 
its return stroke, this cushion steam is compressed isentropically into the 
clearance space; when the piston reaches the end of its stroke, the steam 
has reached the supply pressure at point 5. At point 5, the inlet valve 
opens, and the fresh charge begins to flow into the cylinder; this fresh 
charge is called the flow steam, and its weight will be designated as M /. 
The rest of the cycle is the same as for the reversible adiabatic engine, 
without clearance, when that engine contains, at cutoff, a charge having 
a weight equal to M c + M f . Note that the condition of the steam has not 
changed between points 5 and 1; only the weight is different. The same 
may be said with regard to points 6 and 2. Therefore we may write the 
relation 

Vb _ V* _ M c 
V ! V 2 M c + M f 


( 13 : 2 ) 











THE RECIPROCATING STEAM ENGINE 


309 


and observe that the position of compression (point 6) must, like the posi¬ 
tion of cutoff, be fixed and constant for the reversible adiabatic engine 
with a given clearance. The net work of the reversible cycle is repre¬ 
sented by area 5-1-2-6; this work is performed at the expense of supplying 
a weight M f of fresh charge. The work area equals area 4-1-2-3 less area 
4-5-G-3. Area 4-1-2-3 is the net-work area for the reversible cycle, with¬ 
out clearance, which receives a charge weighing M c + M f and therefore, 
as shown above,- is equal to (M c + M f )(hi — hl) 8 , in Btu. Similarly, 
area 4-5-G-3 may be shown to be equal to ilf c (/i 5 — hl) s or, since the states 
at points 5 and 6 are, respectively, the same as at points 1 and 2, to 
M c (hi — h 2 ) s . Thus the net work of the reversible cycle with clearance is 

= (M c -f- M f) (hi — hi) s — M c (hi — hi) s — M f (hi — hl) s 

rev cycle 

(13:3) 



A comparison of Eq. (13:3) with the equivalent expression for the reversi¬ 
ble engine without clearance shows that the same net amount of work, per 
pound of flow steam , is performed with or without clearance. 


Example 13:2A. Steam is supplied a double-acting reversible adiabatic engine, 
without clearance, at 130 psia dry and saturated. Exhaust pressure is 14.7 psia. 
The engine has a bore of 10 in., a 12-in. stroke, and operates at 180 rpm. Neglect the 
volume of the piston rod. (a) What is the percentage of cutoff? ( b ) What weight 
of flow steam is supplied per cycle? (c) What work is delivered per cycle? ( d ) 
What horsepower does the engine deliver? 

Solution: 

(a) The specific volume of the steam at cutoff (point 1, Fig. 13:2) is that of dry 
and saturated steam at 130 psia, or = 3.455 ft 3 /lb. The specific enthalpy is 
hi = 1191.7 Btu/lb, and Si = 1.5812. Following a line of constant entropy from 
point 1 to a pressure of 14.7 psia on the Mollier diagram, the properties of the steam 
at the end of expansion may be read as h 2 = 1033 Btu/lb and x 2 = 0.878. Then 
v 2 = (0.878) (26.80) = 23.53 ft 3 /lb. The percentage of cutoff is 


Vi = »i 

V 2 Vi 


3.455 

23.53 


= 0.147, or 14.7 per cent 


(6) The volume at cutoff is 


7T (10) 2 (12)(0.147) 
4 1728 


0.08 ft 3 


The weight of flow steam per cycle is 



0.08 

3.455 


0.0232 lb per cycle 


(c) Net work per cycle = M(hi — hl) s = 0.0232(1191.7 — 1033) 

= 3.68 Btu, or 2860 ft-lb, per cycle 


(d) hp = 


(2860) (2) (180) 
33,000 


31.3 hp 






310 


BASIC ENGINEERING THERMODYNAMICS 


Example 13:2 B. The same as Example 13:2 a, except that the reversible adiabatic 
engine has a clearance of 8 per cent at both ends of the cylinder. Calculate (a) per¬ 
centage of cutoff, (6) percentage of compression, (c) weight of cushion steam per cycle, 
(i d ) weight of flow steam per cycle, ( e ) work delivered per cycle, and (/) horsepower 
delivered by the engine. 

Solution: 


(a) Reference is to Fig. 13:3. From Example 13:2A, 



From Eq. (13:1), 

V„ = 0.08(F 2 - Vi) or = 13.5 

Then 


Vi ViV\ 
f 5 v 2 v 5 
v, 

Cutoff = 77 r 


(0.147) (13.5) = 1.98 

F 5 Vx/V h - 1 
f 6 F2/F5 - 1 


1.98 - 1 
13.5 - 1 


0.079, or 7.9 per cent 



F2 _ v 2 

V, ~ 7 l 


1 

0.147 


= 6.8 


„ . F6 — F 5 Vz/V$ — 1 6.8 — 1 „ , 

Compression = 77 - w = tt nr - 7 = To~E- 7 = 0-456, or 45.6 per cent 

V 2 — V 5 r 2 / V 5 — 1 I 0.5 1 

(c) Piston displacement = j = 0.547 ft 3 

4 1 / Zo 

F 6 = (0.456 + 0.08) (0.547) = 0.293 ft 3 
M c = = 0.0109 lb per cycle 

(d) Fi = (0.079 + 0.08) (0.547) = 0.087 ft» 

0 087 

Mf + Me = ^455 = 0.0252 lb per cycle 
Mf = 0.0252 — 0.0109 = 0.0143 lb per cycle • 


(e) Net work per cycle = Mf(hi — h 2 )a — 0.0143(1191.7 — 1033) = 2.27 Btu, or 
1770 ft-lb per cycle 

(1770) (2)(180) 


(/) hp = 


33,000 


19.3 hp 


13:3. The Real-engine Cycle. An indicator diagram such as might be 
obtained during the operation of a real engine is shown by the full lines of 
Fig. 13:4. For purposes of comparison, the dashed lines show a reversible 
cycle having the same clearance volume V 5 and the same volume at cutoff 
Vi. Admission (A) occurs almost at dead center in this real-engine cycle, 
as in the cycle of the reversible engine, but at a pressure somewhat below 
the supply pressure. The cutoff is designated as CO and the lowering of 
line c-CO below the admission line of the reversible cycle is due to the 
differential of pressure necessary to force the steam through the intake 














THE RECIPROCATING STEAM ENGINE 


311 


valve and passages. Because of the lowering of point CO below point 1, 
the expansion line between cutoff and release ( R ) lies below the ideal 
expansion curve. The exhaust valve opens for release before the end of 
the stroke, and the stroke is much shorter than that of the reversible 
engine. This causes the toe-shaped work area of the reversible cycle 
between R and point 2 to be lost, but this area is relatively small and 
could be recovered only at the expense of a greatly lengthened stroke, 
with correspondingly increased losses due to engine friction. The irre¬ 
versible throttling of the steam as it passes through the exhaust valve, 
when that valve is first opened, is shown as line R-d of the real-engine cycle. 
The exhaust line d-K is raised above the line 2-6 by the resistance to flow 
offered by the exhaust valve and passages, and compression (point K) is 



w Fig. 13:4. Comparison of real- and reversible-engine diagrams. 

delayed somewhat, resulting in the lower pressure at admission and lower¬ 
ing the compression curve for the cushion steam. The losses of work area 
due to the lowering of line c-CO below 5-1 and the raising of line d-K above 
2-6 are called wiredrawing losses, and the loss of the toe-shaped area of the 
reversible cycle to the right of R is termed an incomplete-expansion loss. 

A comparison of the areas of the two cycles of Fig. 13:4 does not show 
the real cycle to enclose a much smaller area than that enclosed by the 
reversible cycle, and the thought occurs that the losses due to the various 
irreversibilities are small. This would be true if the weight of flow steam 
were the same for each cycle; this is not the case for, owing to effects to be 
discussed later, the weight of flow steam which must be supplied per cycle 
to the real engine is considerably greater than that necessary for the 
reversible engine. The work measured by the only slightly smaller area 
of the real-engine cycle is therefore credited to a larger weight of steam, 
and the relative performance of the real engine is not as good as would 
appear from a comparison of the areas of Fig. 13:4 alone. 

13:4. Steady-flow Analysis of the Reciprocating Engine. Equation 
(3:5) may be applied to the purpose of analyzing the condition of the 







312 


BASIC ENGINEERING THERMODYNAMICS 


working fluid at exit from the real steam engine. The results of an 
economy test of the engine will show the amount of work performed on 
the piston (from the indicator diagram) per pound of steam supplied. 
Then 


hi + + ~j + iQ 2' — hy + 


TV Z* W e 
2 Jg ^ J ^ J 


in which the subscript 1 refers to the condition of the steam at a section of 
the supply line ahead of the engine and subscript 2' to the state at a sec¬ 
tion of the exhaust line leaving the engine. 1 Q 2 ' is the heat exchange with 
the atmosphere per pound of steam supplied and, being negligible, may be 
assigned a value of zero. W e is obtained, as stated above, from the indi¬ 
cator card, the total indicated work of the engine during the period of the 
test being divided by the total weight of steam supplied during that 
period. The differences between the stored energy due to velocity and 
elevation at entrance and exit are negligible and may be dropped from the 
equation. Then 

W 

h 2 > = hi - (13:4) 


In the usual application of this equation, hi and W e are known from the 
data and results of a test, and hy is the dependent quantity; however, any 
two of the quantities of Eq. (13:4) will suffice to determine the third. 
Thus, if the equation is applied to the ideal reversible engine, h 2 is the 
specific enthalpy corresponding to a state having a pressure equal to that 
at which steam is exhausted from the engine and the same entropy as the 
steam supplied the engine, and W e = J{hi — /i 2 ) s . 

13:5. Engine Efficiency. The term engine is applied in thermo¬ 
dynamics to prime movers, whether they be of the reciprocating or the 
turbine type. It must not be confused with heat engine, which, as we 
have seen, has an altogether different meaning in thermodynamics. The 
efficiency of a heat engine is a ratio of work performance to heat flow, but 
the efficiency of an engine, whether a reciprocating engine or a turbine, is 
the ratio of the work output of the real engine to that of an ideal (reversi¬ 
ble adiabatic) engine to which the working fluid is supplied at the same 
rate and at the same state and which exhausts to the same pressure. In 
the absence of a refrigerator at absolute zero of temperature, it is impossi¬ 
ble for even the ideal heat engine to have an efficiency of 1; on the other 
hand, the perfect prime mover does have that engine efficiency. We may 
write 

W 

v$ = j(h x - h t ). (13 :5) 


where rj e is the engine efficiency, a decimal fraction, W e is the work per- 






THE RECIPROCATING STEAM ENGINE 


313 


formed by the real engine per pound of working fluid supplied to it, and 
(hi — h 2 ) s is the drop in specific enthalpy between the state at which the 
fluid is supplied the real engine and a state having the same entropy but a 
pressure equal to that at which the working fluid is discharged from the 
engine, the difference (hi — h 2 ) s is called the isentropic drop of enthalpy 
and may conveniently be designated as — A h s . The effect of the irreversi¬ 
bilities of the real engine is to decrease the work output below the standard 


set by the reversible adiabatic engine and so, from Eq. (13:4), to raise h 2 
above its value for the ideal engine; this higher value has been designated 
as h 2 > in Eq. (13:4). The indicated work per pound of flow steam supplied 
the real engine is then hi — h v , or 
simply — Ah. In these terms, the 
engine efficiency of the real engine 
becomes 


Ve = 


hi — h 2 ' 


Ah 


(hi — h 2 ) s A h s 


(13:6) 


Equation (13:6) is graphically ex¬ 
plained in Fig. 13:5, a Mollier dia¬ 
gram, in which 1-2 is the process car¬ 
ried out in the ideal engine and 1-2' 
represents the change in state in the 
real engine. A comparison with Fig. 
11:17 and with Eq. (11:27) will show 
that Eq. *(13:6) is as valid for the 
turbine as for the reciprocating engine. 



Fig. 13:5. Engine efficiency. 


In other words, turbine efficiency is essentially the engine efficiency of 
a turbine type of engine. 

The performance of various kinds of prime movers, both vapor and gas, 
is often compared in terms of their respective thermal efficiencies. The 
general definition of the thermal efficiency of an engine is the ratio of the 
work output to the heat supplied from the source during the same time 
period; thermal efficiency is thus equivalent to what has been called in 
these pages the efficiency of a heat engine. Unless the prime mover is a 
complete heat engine in itself, its thermal efficiency cannot always be 
calculated unless the action of other apparatus (such as the boiler of the 
steam power plant) through which the fluid passes is assumed. The 
thermal efficiency of a steam engine is thus really the efficiency of the heat 
engine of which it is a part. In the case of the internal-combustion 
engine, the denominator of the output/input ratio, which is the thermal 
efficiency, is the heat value of the fuel that is furnished the engine during 
the period of time required to deliver the work. Since the fuel is supplied 
directly to the prime mover, the use of the term thermal efficiency is more 












314 


BASIC ENGINEERING THERMODYNAMICS 


reasonable as applied to the internal-combustion engine, even though it is 
not a complete heat engine. 

13:6. Sources of Irreversibility in the Reciprocating Engine. Some of 

the sources of irreversibility in the reciprocating engine have been sug¬ 
gested in Art. 13:3 and pictured in Fig. 13:4. These include the wire¬ 
drawing losses, due to pressure losses through the valves and valve pas¬ 
sages and causing the loss of work area below line 5-1 and above line 2-6 in 
Fig. 13:4, and the incomplete-expansion loss, due to opening the exhaust 
valve before the fluid has dropped to exhaust pressure and causing the loss 
of the toe-shaped work area at the right-hand end of the ideal-engine dia¬ 
gram of that figure. The gain in work area because of the lowering of the 
compression line below that of the ideal diagram is more apparent than 
real since it comes at the expense of delayed compression and thus simply 
means that the weight of cushion steam is less for the real than for the 
ideal cycle. The work expended on the compression of this cushion steam 
is completely recovered by the ideal engine during the performance of its 
succeeding cycle of operation. 

Figure 13:4 does not show nor does Art. 13:3 discuss what is undoubt¬ 
edly the source of the most serious irreversibility, from the standpoint of 
its effect on the efficiency of the reciprocating engine. Although, in the 
course of its traversal of a complete cycle, the reciprocating engine is, very 
closely, an adiabatic device, there is a constant interchange of heat 
between the working fluid and the piston and the walls and head of the 
cylinder. During a part of the cycle this flow of heat is from the fluid into 
these surfaces and, during another part of the cycle, in the opposite direc¬ 
tion. These flows of heat obviously arise from differences of temperature 
between the fluid and the engine parts with which it is in contact; the 
differences of temperature are the result of the intermittent character of 
the fluid flow through the engine. 

The temperature of the fluid varies between extremes set by the tem¬ 
perature at which it is supplied and the lowest temperature that is reached 
in the course of its expansion in the engine. The fresh charge of flow 
steam, as it enters, comes in contact with metal surfaces which have been 
cooled by having been in contact with exhaust steam during at least a 
part of the return stroke of the piston. If the charge is supplied in the 
form of a saturated vapor, a part will be condensed in heating these sur¬ 
faces and will not be available to contribute useful work during the expan¬ 
sion process; if supplied as a superheated vapor, the superheat will at 
least be reduced and some condensation is even likely to take place, with 
a similar effect on work output. This phenomenon is known as initial 
condensation and, in practice, often causes the quality at cutoff to be 0.80 
or less, although dry steam was supplied the engine. The moisture in the 
steam occupies negligible volume, and a correspondingly greater weight 


THE RECIPROCATING STEAM ENGINE 


315 


of vapor is required to fill the cutoff volume (Vi of Fig. 13:4) than for the 
reversible adiabatic engine; thus, for the real reciprocating steam engine, 
the weight of flow steam used per cycle is increased. Moreover, the 
moisture which has accumulated in the charge of the real engine at cutoff 
does not contribute toward the work performed during later expansion to 
release pressure to any appreciable degree. 

As the expansion of the charge proceeds, after cutoff, the temperature 
of the vapor gradually decreases and soon drops below that of the surfaces 
with which it is in contact. The heat exchange is thereafter reversed in 
direction, and, during the latter part of expansion, after release, during 
exhaust and the early part of compression of the cushion steam, heat flows 
from the engine surfaces back into the fluid. This return of heat is called 
reevaporation; it has the effect of slightly lifting the lower part of the 
expansion curve, but the increase of useful-work area is minor and does 
not compare with the loss due to initial condensation. On the thermo¬ 
dynamic basis, the insignificance of work recovery because of reevapora¬ 
tion is partly explained by the lower temperature at which this heat return 
takes place; also, much of the heat is returned during the exhaust stroke, 
with no other effect than to raise the enthalpy of the fluid as it is dis¬ 
charged from the cylinder. The combined effect of initial condensation 
and reevaporation is an irreversibility of major magnitude and importance 
and is called the cylinder condensation loss. 

13:7. Hirn’s analysis is a method of evaluating the heat exchange 
between the fluid and the engine surfaces with which it comes in contact 
during the cycle of the reciprocating engine. To carry out the analysis, a 
typical indicator diagram is needed plus information as to the weight of 
flow steam received per cycle, its state as supplied the engine, the exhaust 
pressure, and the condition of the steam as it was exhausted from the 
engine. The engine bore, stroke, and clearance must also be known. 
The weight of flow steam is usually based on the result of an economy test 
at constant load over a suitable period of time; condenser data taken in 
the course of this test permit the calculation of the exit enthalpy of the 
steam. 

In carrying out the Hirn analysis, the entire cycle is divided into four 
periods, and the heat exchange during each of these periods is separately 
determined. The four periods are: 

1. Admission (starting with the event of admission and concluding at 
cutoff). During this period the fluid system within the cylinder is an 
open system, and Eq. (3:3) may be applied. 

2. Expansion (beginning with cutoff and ending at release). The sys¬ 
tem is closed during this period, and Eq. (2:3) is applicable. 

3. Exhaust (after release and until the event of compression). The sys¬ 
tem is again an open system during this period. Equation (3:3) is applied. 


316 


BASIC ENGINEERING THERMODYNAMICS 


4. Compression (between the events of compression and admission). 
Again the system is a closed system, and Eq. (2:3) may be applied. 

As it applies to the admission period, Eq. (3:3) may be written (neglect¬ 
ing energy quantities due to velocity and elevation as insignificant) in the 
form 

(Me + M,)uco - M c u a = - + a Qco + MM 

in which the subscripts A and CO refer to the events of admission and 
cutoff, respectively, and the subscript 1 to the state at which steam is sup¬ 
plied the engine; M c and Mf are the weights of cushion and of flow steam 
per cycle, respectively; aW co is the work performed on the piston and may 
be obtained as the area between the line connecting admission and cutoff 
on the indicator diagram and the V axis of that diagram, expressed in 
foot-pounds; and A Qco is the heat exchange with the cylinder and piston 
during the admission period, a negative sign indicating heat flow from the 
charge of steam into the walls. The state of the fluid corresponding to 
any point on the cycle of the indicator diagram can be determined from 
the pressure and volume at that point, as measured from the indicator 
diagram, if the weight of fluid present in the cylinder is known. For 
example, at cutoff this weight is Mf + M c ; of these two weight compo¬ 
nents, M f is known from data secured before the Hirn analysis is applied. 
M c , the weight of cushion steam, is conventionally calculated on the basis 
of the assumption that, at the point of compression (K of Fig. 13:4), the 
steam has a quality of 1; this is usually not far from the actual quality 
because of the effects of reevaporation and drainage of the condensate, 
but, in any event, the assumption does not lead to serious error because of 
the relatively small weight of cushion steam as compared with the flow 
steam. All terms of the equation may thus be separately determined 
except the heat flow during the admission period. Rearranging the equa¬ 
tion into the form 

aQco = (M c + Mf)uco — M c ua — M /hi T A - j C - (13:7) 

this heat flow may be computed. 

Applying Eq. (2:3) to the expansion period, 

coQk = ( M c + Mf) (u R — Uco) + — j— (13:8) 


in which the terms on the right-hand side of the equation are separately 
evaluated in a manner identical with that used in the case of the similar 
terms of Eq. (13:7). 

Similarly, applying Eq. (3:3) to the exhaust period, we may write 


rQk — M C U K -f- Mfh* — (M c -f- Mf)u R -f- 




J 


(13:9) 






THE RECIPROCATING STEAM ENGINE 


317 


in which h 2 > is the specific enthalpy of the exhaust steam and may be deter¬ 
mined from test data taken from the condenser. 

Finally, the application of Eq. (2:3) to the compression period gives 

*AV A 

kQa — M c (u a — Uk) H- -j— (13:10) 


The algebraic sum of the four individual heat flows should be found to 
be very nearly equal to zero. The difference from zero, if any, should be 
negative and represents heat transferred through the walls and head of 
the cylinder and into the surrounding atmosphere. In the absence of 
complete condenser data, it is sometimes necessary to fix the value of h 2 > 
by applying Eq. (13:4); this amounts to an assumption that the real 
engine is an adiabatic (though not a reversible) engine and assures that 
the sum of the four individual heat flows will be zero. 


Example 13:7. A steam-engine economy test supplies the following data: 


Engine data: Double-acting, 10-in. bore, 12-in. stroke, 180 rpm. Neglect piston-rod 
volume. 

Steam data: Supply pressure, 130 psia, dry and saturated. 

Condenser data: Exhaust pressure, 14.7 psia. Condensate temperature, 200°F. 
Weights, per hour: condensate, 1062 lb; cooling water, 31,800 lb. Temperatures, 
cooling water: inlet, 65°F; outlet, 96°F. 

Indicator-card data: Scale of indicator spring, 80 psi/in. Length of indicator card, 
4 in. Location of events: admission, 0 per cent, 90 psia; cutoff, 20 per cent, 115 
psia; release, 95 per cent, 31 psia; compression, 30 per cent, 18 psia. 

Indicator-card areas (measured to V axis at p = 0 psia): 


Admission period (A to CO) = 
Expansion period {CO to R) = 
Exhaust period (R to K ) = 

Compression period (if to 4) = 

Net (enclosed) area 


1.26 in. 2 
2.24 in. 2 
-0.54 in. 2 
-0.66 in. 2 


. = 2.30 in. 2 

Calculate (a) work per period per cycle and net work of the cycle, Btu; (6) indicated 
horsepower of the engine; (c) weights of cushion steam and flow steam per cycle; 
{d) quality at cutoff, release, and admission; ( e ) heat exchange with the walls during 
each of the four periods; (/) heat flow to the atmosphere per cycle; ( g) h 2 ' as for an 
adiabatic engine; ( h ) engine efficiency based on the indicated horsepower. 

Solution: 


{a) Piston displacement = 0.547 ft :i (see Example 13:2B, part c). 

, (80) (144) (0.547) 0 no 

of indicator-card area = - (778) (4) - “ mu ' 


One square inch 


aWco 

J 

CoW R 


= 1.26(2.02) = 
=* 2.24(2.02) - 


2.54 Btu 
4.52 Btu 







318 


BASIC ENGINEERING THERMODYNAMICS 


rWk 

J 


= -0.54(2.02) 


— 1.09 Btu 


kWa 

J 

Net work 


-0.66(2.02) = -1.33 Btu 
= 4.65 Btu 


(b) ihp = 


(4.65) (180) (2) (778) 
33,000 


39.4 hp 


(c) V A = (0.08) (0.547) = 0.0438 ft 3 

Vco = (0.08 + 0.20) (0.547) = 0.1535 ft 3 

V R = (0.08 + 0.95) (0.547) = 0.5630 ft 3 

V K = (0.08 + 0.30) (0.547) = 0.2080 ft 3 


The quality at compression is assumed to be 1. Then 


M c = 
M f = 
(d) xco : 


V K 


v g at 18 psia 
1062 

(180) (2) (60) 


0.2080 

22.17 


= 0.0094 lb 


= 0.0492 lb per cycle 
Vco 


0.1535 


XR = 


{Me + Mf){v g at 115 psia) (0.0094 + 0.0492)(3.882) 

Vr 0.5630 A _ 0 

= 0./2 


= 0.675 


(Me + M f )(v g at 31 psia) (0.0586) (13.33) 

= Va = °- 0438 = 0952 

M c (v„ at 90 psia) (0.0094) (4.896) 

xr = 1-00 (assumed) 

(e) u A = 290.3 + (0.952) (813.4) = 1064.7 Btu/lb 
uco = 308.7 + (0.675) (798.4) = 847.8 Btu/lb 
u R = 220.6 + (0.72) (867.7) = 845.3 Btu/lb 
uk = 1080.4 Btu/lb; hi = 1191.7 Btu/lb 

hi: = enthalpy of condensate + heat removed by cooling water per pound of 

steam condensed 

= 168.0 + 31,8QQ(63 1 ^ 2 ~ - 33, ° 5) = 168 + 926 = 1094 Btu/lb 

aQco = (0.0586)(847.8) - (0.0094)(1064.7) - (0.0492)(1191.7) + 2.54 

= 49.70 - 10.00 - 58.70 + 2.54 = -16.46 Btu [Eq. (13:7)] 

coQr = 0.0586(845.3 - 847.8) + 4.52 = -0.15 + 4.52 = +4.37 Btu [Eq. (13:8)] 
rQk = (0.0094) (1080.4) + (0.0492) (1094) - (0.0586) (845.3) - 1.09 

= 10.15 + 53.85 - 49.50 - 1.09 = +13.41 Btu [Eq. (13:9)] 

kQa = 0.0094(1064.7 - 1080.4) - 1.33 = -0.15 - 1.33 = -1.48 Btu 

[Eq. (13:10)] 

(/) Qcycie = -16.46 + 4.37 + 13.41 - 1.48 = -0.16 Btu 


As indicated by its negative sign, this is the heat flow to the atmosphere. As a check, 
see the calculation to follow. 

(g) Work per pound of steam = — =-— ^^ — = Q4Q2 = 94.5 Btu/lb 

From Eq. (13:4), 

hn = 1191.7 - 94.5 = 1097.2 Btu/lb 

Note that this value of h? assumes the engine to be an adiabatic engine. The differ¬ 
ence between the two values of h 2 > is 1094 - 1097.2 = —3.2 Btu/lb. Per cycle, this 


















THE RECIPROCATING STEAM ENGINE 319 


difference is (0.0492)(—3.2) — —0.16 Btu. Considering that all calculations were 
made on the slide rule, this check is much closer than was to be expected. 


(h) 


We 

J (hi — h 2 )s 


or 


94.5 

1191.7 - 1033 


0.595 


Ve 


hi — h 2 > 
(hi — h 2 ) s 


1191.7 - 1097.2 
1191.7 - 1033 


0.595 


[Eq. (13:5)] 


[Eq. (13:6)] 


Note that here it was necessary to use h 2 > as calculated for an adiabatic engine. Equa¬ 
tion (13:5) is the basic expression for engine efficiency; Eq. (13:6) is valid only for an 
adiabatic (though not necessarily a reversible) engine. 


13:8. Reduction of Cylinder Condensation. It is evident that the 
loss due to cylinder condensation will vary in magnitude with the mass of 
metal that is successively heated and cooled per pound of flow steam per 
cycle and with the extreme variation of temperature of this metal in the 
course of the cycle. It follows that any steps that are taken with a view 
toward reducing the effects of cylinder condensation on engine efficiency 
should operate either to reduce the mass of metal per pound of flow steam 
or to reduce the interval through which the temperature of the cylinder 
fluctuates. 

The temperature of the cylinder will always lag behind the temperature 
of the steam it contains so that, while the engine is in operation, the 
highest temperature reached by the cylinder will never be as high as the 
temperature"of the steam as supplied the engine nor will the cylinder tem¬ 
perature fall as low as that of the exhaust steam. Increasing the differ¬ 
ence between the ranges of temperature through which the steam and the 
metal of the cylinder and piston fluctuate will have a beneficial effect in 
reducing cylinder condensation losses. Two general methods of lowering 
the ratio which the extreme fluctuation of cylinder temperature bears to 
the difference between the limiting temperatures of the steam in the course 
of the cycle are: 

1. Decreasing the time period during which the metal surfaces are 
exposed to steam at the highest and lowest temperatures of the cycle. 

2. Reducing the time rate of heat transfer between steam and metal. 
The first method may be implemented by either shortening the period of 
admission and/or the period of exhaust in the cycle to a smaller fraction 
of the complete cycle or by increasing the speed with which the entire 
cycle is traversed. With reference to the second method, it has been 
found that the rate of heat transfer is increased in proportion to the 
wetness of the cylinder walls and this, in turn, will depend on the per¬ 
centage of moisture in the steam contained within the cylinder; by supply¬ 
ing steam to the engine highly superheated, the time rate of heat transfer 
can thus be reduced. 






320 


BASIC ENGINEERING THERMODYNAMICS 


When cutoff takes place at a later point in the power stroke, it might 
appear from the preceding paragraph that the effect would be to increase 
the bad effects of cylinder condensation because of the lengthening of the 
period of admission. This is balanced by the opposing argument that the 
weight of flow steam per cycle is greater for longer cutoffs and the total 
effect of the fluctuation of cylinder temperature is less per pound of flow 
steam. Up to a certain limit, therefore, long cutoffs are beneficial in 
reducing the effects of cylinder condensation on engine efficiency; this 
limit is often set by the introduction of increasing incomplete-expansion 
losses by delaying the cutoff. 

Some specific measures which have been taken in engine design in the 
effort to reduce cylinder condensation losses include: 

1. High speed. While engine speeds do not compare with turbine 
speeds, there is a considerable variation according to the size of the engine 
and type of valve used. 

2. Use of superheated steam. The effect of supplying steam to the 
engine as superheated steam has been discussed above. The correspond¬ 
ingly higher steam temperatures sometimes require the use of special 
valves, designed to resist warping. 

3. Corliss valves. The engine of Fig. 13:1 has valves of the Corliss 
type. This design provides separate ports for incoming and outgoing 
steam and reduces the mass of the metal that is alternately heated and 
cooled in the course of the cycle. Also, the ports are much shorter than 
in the slide-valve engine, with the same effect in reducing the ratio of the 
mass of metal to the mass of flow steam. 

4. Multiple expansion. To carry out the total expansion of the steam 
in stages, as is so commonly done in the turbine, requires, in the recipro¬ 
cating engine, a separate cylinder for each stage of the expansion, with 
correspondingly increased losses due to mechanical friction; therefore the 
multiple-expansion reciprocating engine is usually built as a compound 
(two-stage-expansion) engine, although triple- and quadruple-expansion 
engines are in use. In single-expansion, there are two lags, or intervals, 
between the temperature of the steam and the temperature of the cylin¬ 
der; one of these is the temperature interval between the highest tempera¬ 
ture at which steam is supplied the cylinder and the somewhat lower peak 
temperature reached by the metal surfaces; the other is the difference 
between the temperature of the exhaust steam and the lowest temperature 
of the cylinder. In the compound engine, there are four such lags, two 
for each stage of expansion, and the total range of the fluctuation of cylin¬ 
der temperature is somewhat lessened. Moreover, the high-pressure 
stage of the expansion is carried out in a smaller cylinder, thus reducing 
the mass of metal per pound of flow steam per cycle in that stage. 

5. Uniflow design. The possibility of moving the exhaust valves from 


THE RECIPROCATING STEAM ENGINE 


321 


the head of the cylinder has been mentioned previously. In the uniflow 
engine the exhaust valves are mere slots in the cylinder wall, being placed 
near the outer end of the piston stroke. These slots are uncovered by the 
piston near the end of the power stroke, allowing a major portion of the 
steam to escape from the cylinder; as the piston begins its return stroke, 
the slots are again closed and the remainder of the charge is compressed. 
As has been pointed out above, the greater amount of work thus expended 
on compression is recoverable during the succeeding cycle. A long piston 
makes it possible to use the same slots, placed at the center of the cylinder, 
for the exhaust from both ends in double-acting design. The name uni- 
flow derives from the fact that the steam enters at one end of the piston 
stroke and leaves at, or near, the other, thus giving one-way flow; when 
other types of valves are used, the term counterflow is employed to describe 
the flow of steam through the cylinder. The advantage of the uniflow 
engine in reducing the effects of cylinder condensation is due to its reduc¬ 
tion of the total interval through which the temperature of the cylinder 
fluctuates; this comes as the result of the shortening of the period of 
exhaust. This advantage is increased because exhaust steam never flows 
over the surfaces near the admission end of the cylinder and initial con¬ 
densation effects are especially reduced, thus increasing the quality at 
cutoff. Uniflow design is quite effective, producing steam rates (pounds 
of steam used per horsepower-hour or per kilowatt-hour) which are com¬ 
parable with those obtained by compounding; the uniflow engine is less 
bulky and less costly than the compound engine. 

13:9. Methods of Governing the Engine. Both the reciprocating 
engine and the turbine are usually designed to operate at constant speed 
under variable load. To adjust the output of the engine to the fluctuating 
load, some means of controlling the rate at which steam is supplied the 
engine must be provided; this control is called governing. 

There are two basic methods of governing an engine or turbine, as 
already suggested in Art. 11:12 with reference to the governing of the 
turbine. The first of these is cutoff governing. Cutoff governing, as 
applied to the reciprocating engine, consists in changing the position of 
cutoff to conform to the load on the engine. This requires the provision 
of a valve mechanism so designed that the position of cutoff may be 
altered, usually automatically, while the engine is in operation. Although 
the engine efficiency, may be slightly changed, principally owing to the 
effect of a change in the valve setting on the incomplete-expansion and 
cylinder condensation losses, steam is supplied the cylinder at unchanged 
pressure and the exhaust pressure also remains the same. Thus the 
maximum work that could be obtained from each pound of flow steam in 
an equivalent reversible adiabatic engine is not affected. 

The second method is called throttle governing and consists in control of 


322 


BASIC ENGINEERING THERMODYNAMICS 


engine output by throttling the steam to lower pressure before it enters 
the steam chest of the reciprocating engine; the position of the valve 
events is not changed. This has the effect of lowering the top line of the 
indicator diagram and so reducing the net-work area enclosed. As shown 
in Fig. 13:6, this method of governing, although it can be used with a 
valve of simpler and less costly design, is thermodynamically expensive. 
In the figure, steam is supplied the engine by the boiler at the state which 
is designated by point 1; as the result of an isentropic expansion to exhaust 
pressure p 2 , the reversible adiabatic engine could deliver the difference 
hi — h 2 , represented by the length of the line 1-2 in the figure, in the form 

of work. Under lighter load, the gov¬ 
ernor steps in to throttle the steam to a 
lower pressure p x before it enters the 
cylinder. This does not change the 
enthalpy of the steam, as we have seen 
in Chap. 7, but does increase its en¬ 
tropy, and the new supply state is indi¬ 
cated by point 3 of Fig. 13:6. A re¬ 
versible adiabatic engine, accepting the 
steam at state 3, could discharge it at 
the state corresponding to point 4 of the 
figure, delivering the enthalpy differ¬ 
ence hz — hi in the form of work. This 
is a less effective use of the energy in 
the steam than the expansion along 
line 1-2, as is evident from the figure. 
Cutoff governing is thus thermody¬ 
namically superior to throttle governing; the same conclusion may, 
of course, be applied to the governing of turbines. 

13:10. Reciprocating Engine versus Turbine. The reversible-adia¬ 
batic-turbine and the reversible-adiabatic-reciprocating-engine prime 
movers are alike in the amount of work they can, in theory, deliver under 
parallel conditions of steam supply and exhaust pressure. The choice 
between the two types of prime mover must be based on practical factors 
such as cost, maintenance, adaptability to the conditions under which 
power is to be delivered, space available, and the percentage of ideal per¬ 
formance that can be realized (the engine or turbine efficiency); this 
last-named factor will depend on the relative total magnitude of losses 
such as have been listed for the two types of prime mover. 

The turbine must operate at high rotative speeds for high efficiency. 
4 his high speed keeps its size, weight, and cost for a given rated power 
much lower than those of the reciprocating engine and makes it possible 
to build the turbine in units of a capacity more than ten times greater than 



5 


Fig. 13:6. Thermodynamic effect of 
throttle governing. 







THE RECIPROCATING STEAM ENGINE 


323 


that of the largest practical reciprocating engine. On the other hand, 
turbines are wasteful of steam in situations where a high starting torque 
is required and do not retain their efficiency as well as does the recipro¬ 
cating engine under variable-speed and variable-load conditions. The 
direction of rotation of the turbine cannot be reversed as is possible for 
the reciprocating engine in some of its forms. 

The thermodynamic losses associated with the reciprocating engine are 
those due to wiredrawing, to cylinder condensation, to incomplete expan¬ 
sion, and to mechanical friction. In the turbine the flow of steam is 
steady and continuous and there is, under steady-load conditions, no 
intermittent opening and closing of valves as in the reciprocating engine; 
the turbine has no loss comparable with the wiredrawing loss of the engine. 
Under steady-load conditions, the pressure and temperature at any given 
point in the steam turbine remain constant; thus there is no fluctuation of 
temperature and no cylinder condensation loss. The factors which, in 
the engine, often make it desirable to open the exhaust valve before 
exhaust pressure is reached by the expanding charge are not present in 
the turbine; the turbine has no incofriplete-expansion loss. Finally, in 
the turbine there is no need to transform reciprocating into rotary motion; 
the loss due to mechanical friction is much smaller for the turbine even 
when gears must be used to reduce its speed. 

But the turbine has losses of its own which either are not experienced 
in engine operation or are much less important in their effect on engine 
efficiency. These have been discussed in Chap. 11, and the more impor¬ 
tant may be summarized as: 

1. Higher fluid turbulence losses, due to very much greater fluid 
velocities. 

2. Rotation losses, due to rapid movement of blades and rotors. 

3. Residual velocity losses, due to unremoved kinetic energy in the 
steam at exit from the turbine. 

4. Leakage around turbine blades, due to clearance between the rotating 
blades and the turbine casing. 

The effect of the characteristic losses associated with the turbine 
amounts, in total, to much the same deduction from ideal performance as 
the typical reciprocating-engine losses cause in reducing the performance 
of the engine. Turbine and reciprocating-engine efficiencies are of the 
same order and depend primarily on refinement of design. The care and 
expense which are justified in reducing losses are, in general, a function of 
the rated power for either type of prime mover. The maximum size in 
which it is practical to build the reciprocating engine is about 10,000 kw, 
and a representative efficiency for a reciprocating engine of this capacity 
is of the order of 80 per cent; in general, about the same efficiency would 
apply to a turbine unit of 100,000 kw. 


324 


BASIC ENGINEERING THERMODYNAMICS 


Problems 

1. Work Example 13:2 A, changing the exhaust pressure to 10 psia. 

2. Work Example 13:25, changing the exhaust pressure to 10 psia. 

3. Work Example 13:25, changing the supply pressure to 200 psia. 

4. Work Example 13:25, changing the clearance to 6 per cent. 

5. During a test which is made at constant load, a double-acting engine operates 

at 140 rpm, receiving saturated steam at 200 psig and exhausting to the atmosphere 
(p = 14.7 psia). The average indicator card has an area of 3.4 in. 2 , the indicator 
spring having a scale (the scale of indicator-card ordinates) of 100 psi/in., and the 

scale of indicator-card abscissas being 0.5 ft 3 /in. The test shows that 1200 lb of 

steam is supplied the engine during a 15-min period, (a) What is the indicated 
horsepower at which the test was conducted? (6) What is the engine efficiency 
based on the indicated horsepower? (c) What is the quality of the steam as it is 
exhausted from the engine? 

6. An engine receives 1000 lb of steam at 240 psia, 98 per cent quality, during a 
20-min constant-load test. The condition of the steam in the exhaust line is p = 4 
psia, x = 0.90. Assume an adiabatic engine, (a) What indicated horsepower does 
the engine develop? ( b ) What is the engine efficiency? 

7. Steam is supplied to an engine at 200 psig as saturated steam. At cutoff, which 
takes place at 25 per cent of the stroke,'the quality is 77 per cent because of initial 
condensation; the pressure at cutoff is 185 psig, owing to wiredrawing. The piston 
displacement of the engine is 2 ft 3 , and the clearance is 6 per cent. What weight of 
H 2 0 is present in the cylinder at cutoff? If release takes place at 95 per cent of the 
stroke and the pressure is 42 psig at release, what is the quality of the steam at 
release? 

8. In Prob. 7, the event of compression is located at 30 per cent of the stroke as 
read from the indicator card, and the pressure of the steam at that point is scaled from 
the indicator card as 2 psig. Assuming the steam to be dry at release, what is the 
weight of cushion steam? The weight of flow steam? 

9. In engine design, the designer often approaches the problem by laying out an 
ideal, or conventional, indicator diagram which is based on the pressures at which 
steam is supplied and exhausted, a planned cutoff percentage at full load (usually 
40 per cent or less), and zero clearance. This ideal card has an appearance similar to 
Fig. 13:2 except that, since the cutoff percentage has been independently selected, 
expansion will not be complete and the pressure at the end of the expansion curve 
which begins at cutoff will be higher than the exhaust pressure. Release is assumed 
to be located at the end of the power stroke, and the pressure at release is calculated 
on the assumption that the product PV, where P is the absolute pressure, is constant 
between cutoff and release. Following release, the ideal diagram shows a vertical 
drop of pressure to exhaust pressure and, following that, a rejection of steam from the 
cylinder at constant pressure as in Fig. 13:2. Sketch an ideal indicator card, based 
on supplying saturated steam at 150 psia, exhausting at 3 psia, with cutoff assumed to 
be at 30 per cent. What is the pressure at release? What is the mean effective 
pressure shown by this ideal indicator card? 

10. Having constructed an ideal diagram as in Prob. 9, the designer applies a 
diagram factor which his experience tells him will approximately express the ratio of 
the mean effective pressure of the real indicator diagram taken from an engine operat¬ 
ing under equivalent conditions to the mep of the ideal diagram. Assuming a diagram 
factor of 0.80, calculate the indicated horsepower which will be developed by a simple 
double-acting steam engine of 15 in. bore and 20-in. stroke, with 8 per cent clearance, 


THE RECIPROCATING STEAM ENGINE 


325 


operating at 120 rpm. The steam supply and exhaust pressures and the percentage 
of cutoff are as in Prob. 9. 

11. In Prob. 10, assume that the pressure at cutoff is 140 psia and the quality of the 
steam at that point in the stroke is 0.75. Also assume that compression is at 25 
per cent, the pressure at compression is 4 psia, and the steam remaining in the cylinder 
at compression is dry. What weight of steam is supplied the engine per hour? What 
is its steam rate (pounds of steam supplied the engine per indicated horsepower-hour 
delivered)? What is the engine efficiency? 

12. An engine is double-acting, 12 in. bore and 15-in. stroke, and operates at 150 
rpm. Its clearance is 8 per cent at each end of the cylinder. During a 30-min test 
at constant load, it is supplied with 1000 lb of saturated steam at 140 psia. The 
condenser pressure is 14.7 psia. Condensate leaves the condenser at 200°F. Per 
pound of steam condensed, 35 lb of cooling water passes through the condenser, 
entering at an average temperature of 70°F and leaving at an average temperature of 
96.4°F. Events as shown on the indicator card occur as follows: 

Cutoff at 25 per cent, 130 psia. 

Release at 95 per cent, 40 psia. 

Compression at 30 per cent, 18 psia. 

Admission at 0 per cent, 85 psia. 

The scale of the indicator spring (the scale of ordinates on the indicator diagram) is 
80 psi/in. The length of the indicator diagram (representing the total piston dis¬ 
placement) is 4 in. Areas are measured from the card beneath the lines connecting 
successive events and above a horizontal line representing the pressure of the atmos¬ 
phere (14.7 psia) and found to be as listed below. A positive sign indicates that work 
is done by the steam. 

Admission to cutoff, +1.50 in. 2 
Cutoff to release, +1.90 in. 2 
Release to compression, —0.10 in. 2 
Compression to admission, —0.31 in. 2 

What are (a) the weight of cushion steam, based on the usual assumption, ( b ) the weight 
of flow steam, and ( c ) the quality at cutoff? Calculate the amount of heat transfer, 
and state the direction of its flow between ( d ) admission and cutoff, (e) cutoff and 
release, (/) release and compression, and ( g ) compression and admission, (h) What 
is the net heat transfer for the cycle? Is this an adiabatic engine? (i) What is the 
efficiency of the engine? (j) What is the steam rate (pounds of steam supplied per 
horsepower-hour delivered)? 

Symbols 

g acceleration of gravity 
h specific enthalpy 
J proportionality constant 
M mass of a system 
p pressure, psi 

P pressure, psf; pressure in general 
Q rate of heat flow 
s specific entropy 
u specific internal energy 
v specific volume 

V volume; volume of a system 

V velocity 

W rate of work delivery per cycle or per unit time 


326 


BASIC ENGINEERING THERMODYNAMICS 


x quality of wet steam 
z elevation 

Greek Letters 
rje engine efficiency 

Subscripts 

A admission 
c cushion steam 
CO cutoff 
D displacement 
e engine 
/ flow steam 
K compression 
R release 
rei 7- reversible 

s constant entropy 


CHAPTER 14 

POWER—VAPOR SYSTEMS 


14.1. The Carnot Cycle as a Vapor Cycle. It will be remembered that 
the Carnot cycle requires that heat be supplied at the constant tempera¬ 
ture of the source to the working fluid and that the rejection of heat must 
be at the constant temperature of the refrigerator. Heat exchange at 
constant temperature has not proved practical when the working sub¬ 
stance is a gas or a superheated vapor, 
and wide variations from the ideal cycle 
have resulted. But a heat exchange 
at constant temperature also takes 
place at constant pressure if the fluid 
is a saturated vapor. In a steady-flow 
heat exchanger, such as a boiler or a 
condenser, the pressure of the fluid 
remains essentially constant as it passes 
through the unit. 1 Thus one of the 
objections to the Carnot cycle as a 
practical cycle is removed. Further, 
as a steady-flow cycle, it is no longer 

necessary that the working cylinder 

,iv, v ,v , r Fig. 14:1. The Carnot vapor cycle, 

act as the heat exchanger; that func¬ 
tion can be transferred to devices that are better adapted for the purpose. 

In Fig. 14:1, a Ts diagram, a Carnot cycle has been placed between the 
saturated liquid and dry-saturated-vapor lines of the vapor which is the 
working substance. The process 1-2 represents the addition of heat to 
the vapor; this is carried out in a boiler which, in the cycle illustrated, 
receives the fluid as a saturated liquid at the temperature T s and delivers 
it at the same temperature (and with no change in pressure) but as a dry 
and saturated vapor. Since the boiler, as a practical device, operates at 
essentially constant pressure, line 1-2 is shown in its greatest possible 
length; in the figure, if point 1 were moved to the left and/or point 2 to 

the right, the pressures would differ at supply and discharge. The 

* 

1 Because of its viscosity, the real fluid will always show at least a small drop in 
pressure through the steady-flow heat exchanger, but as rates of flow (velocities) 
decrease, this differential of pressure necessary to overcome flow resistance approaches 
zero as a limit. 



327 










328 


BASIC ENGINEERING THERMODYNAMICS 


amount of heat supplied per pound of fluid flow is represented by the area 
1-2 -b-a. This heat is supplied during steady flow through the boiler and, 
neglecting differences in stored energy due to elevation and velocity at 
entrance to and exit from this unit, may be shown, according to Eq. (3:5), 
to be equal to the difference of enthalpy at the two stations, h 2 — hi. 

The rejection of heat represented by process 3-4 takes place in a steady- 
flow heat exchanger called a condenser and is carried out at the constant 
temperature T R of the refrigerator. Since points 3 and 4 both lie within 
the saturated-vapor region, this heat is rejected at constant pressure, a 
practical method of steady-flow heat exchange. The heat rejected per 
pound of fluid flow in the condenser is represented by area 4-3 -b-a of the 



figure and, again neglecting differences of elevation and velocity at 
entrance and exit and applying Eq. (3:5), is equal to h 3 — h 4 . 

The isentropic expansion process 2-3 is carried out in a reversible 
adiabatic prime mover which delivers work to external systems in the 
amount h 2 — h 3 per pound of fluid. A part of this work is returned to 
drive the reversible adiabatic compressor that is required to close the 
cycle by carrying out process 4-1; the amount of work so returned is 
hi — The flow diagram of the Carnot steady-flow vapor cycle is 
shown in Fig. 14:2. 

Although an examination of the proportions of Fig. 14:1 will show that 
the efficiency of the Carnot steady-flow vapor cycle is (T s — T R )/T S and 
thus the maximum thermodynamically conceivable for a heat engine and 
although many of the practical difficulties that block the demonstration 
of the Carnot nonflow cycle have been avoided, the Carnot vapor cycle is 
still not a practical cycle and is not used as the basis of a real power plant. 
This is largely because the vapor-compression process 4-1 is difficult to 
accomplish. This process would require that the fluid be removed from 















POWER—VAPOR SYSTEMS 


329 


the condenser before condensation had been completed; because there is 
no indication in terms of changes in temperature or pressure as to when 
point 4 has been reached and because, owing to the great difference in 
their densities, the liquid and the vapor portions of the mixture tend to 
separate during condensation, the removal of the vapor as a homogeneous 
mixture at condition 4 is not easily managed. Again, although the ratio 
of the net work of the cycle to the gross work of expansion is more favor¬ 
able than when the working fluid is a gas, this ratio still suffers by com¬ 
parison with the corresponding ratio of vapor cycles to be discussed in 
following pages of this chapter. The effect of a low work ratio in making 
a cycle more vulnerable to the irreversibilities of the real engine has been 
pointed out in Example 4:9. 

Example 14:1. (a) What is the efficiency of a Carnot vapor cycle in which steam 

at 600 psia is supplied the prime mover and the condenser pressure is 1 psia? (6) If 
the fluid is to enter the boiler as a saturated liquid, at what quality must it be with¬ 
drawn from the condenser? ( c) Assuming that the steam is dry as it enters the prime 
mover, what amount of work is performed in that unit per pound of steam flow? 
(d) What amount of work must be returned for compression? (e) What are the net 
work of the cycle and its ratio to the gross work of expansion? 

Solution: 

(a) The saturation temperatures of steam at 600 and 1 psia are, respectively, 486.21 
and 101.74°F. Then 


v 


486.21 --101.74 
486.21 + 460 


0.407 


(6) S4 = Sl = 0.6720 = s/ 4 + XtSf^ = 0.1326 + x 4 (1.8426) or rr 4 = 0.292 

(c) Reading from the Mollier diagram, h 2 = 1203 Btu and h% = 807 Btu. Then 


= h 2 — h 3 = 1203 - 807 = 396 Btu/lb 

J 

(< d) h A = h fi + x 4 h f0i = 69.7 + (0.292)(1036.2) = 372.2 Btu; hi = 471.6 Btu 
.Ef = hi - h x = 372.2 - 471.6 = -99.4 Btu/lb 

J 

The negative sign indicates that this work is returned. Note that, having determined 
the enthalpies hi, h 2 , h h and hi, the efficiency as calculated in part a may be checked, 
as follows: 

Qs - Qr {h 2 - hi) - (h s - hi) _ (1203 - 471.6) - (8 07 - 372.2) 

11 = ~~Q~s h 2 - hi 1203 - 471.6 

= 0.407 

( e) Net work of cycle = 4— -j~ = 396 — 99.4 = 296.6 Btu/lb 

Work ratio = = 0.75 

396 







330 


BASIC ENGINEERING THERMODYNAMICS 


This answer may be compared with the work ratio calculated in part d of Example 4:9 
for a Carnot cycle employing a gas as the working substance. Although the tempera¬ 
tures of source and refrigerator are nearly the same in the two examples, the work 
ratio of the vapor cycle is much more favorable. 

14:2. The Rankine cycle is illustrated 
on Ts coordinates in Fig. 14:3. In the 
Rankine cycle, concessions are made to 
practicability by specifying that the 
transfers of heat that take place in the 
vapor generator (the boiler and, some¬ 
times, the superheater) and the con¬ 
denser shall be at constant pressure and 
that the heat-rejection process in the 
condenser shall continue until conden¬ 
sation is completed, i.e., the fluid exits 
from the condenser as a saturated liquid. 
For the limiting temperatures T s and 
T r of the Carnot cycle are substituted 
limiting pressures p u and pi , respectively, 
the upper and lower pressures reached by the fluid in its passage through 
the cycle. The flow diagram of the Rankine-cycle plant is shown in Fig. 
14:4. The boiler receives the fluid as a compressed liquid at the pressure 
p u . As heat is added in that unit, the first effect on the condition of the 



Fig. 14:3. The Rankine cycle. 



fluid is to raise the temperature of the compressed liquid (process 1-c of 
Fig. 14:3) until at c it has reached the saturation temperature equivalent 
to the pressure p u at which the boiler operates. As further heat is added 
(process c- 2), still within the boiler, the liquid evaporates at constant 
pressure p u (and constant temperature) until it leaves the boiler as a dry 























POWER—VAPOR SYSTEMS 


331 


saturated vapor at condition 2. The operation that takes place in the 
boiler is essentially a steady-flow process; the differences in stored energy 
of the fluid due to differences of elevation and velocity at entrance to and 
exit fiom the boiler may be considered negligible, and no external work 
accompanies the process. Applying Eq. (3:5), it may be shown that 

Qs — h 2 — hi (14:1) 

where Q s = heat supplied per pound 

hi = specific enthalpy of compressed liquid at entrance to boiler 
h 2 = specific enthalpy of the vapor as it leaves boiler 
Process 2-3 is carried out in a reversible adiabatic prime mover which is 
identical in type with the corresponding unit employed in the Carnot 
cycle. As shown in Chap. 11 and 13, the work delivery per pound of fluid 
flow is 

W 

-f = h t ~ hz (14:2) 

in which ha is the specific enthalpy of the fluid at the same entropy as when 
it entered the prime mover but at the lower pressure of the cycle, pi. 

The condenser carries out process 3-4, removing heat from the exhaust 
vapor until it leaves that unit as a saturated liquid at the lower pressure 
of the cycle, p t . The specific enthalpy at exit, hi, may thus be designated, 
for clarity, as h fr Again the differences in elevation and velocity at 
entrance and exit from the condenser are inconsiderable, and there is no 
external work. Applying Eq. (3:5) to this steady-flow heat exchanger, 
the heat rejected per pound of fluid is observed to be 

Qr — ha hi = h 3 — h/ t (14:3) 

in which the use of the subscripts/and l indicates that hi is the enthalpy of 
the saturated liquid at the lower pressure of the cycle. 

In place of the vapor compressor of the Carnot vapor cycle, the Rankine 
cycle substitutes a liquid pump (called a feedwater pump in the steam 
power plant) which receives the fluid as a saturated liquid and compresses 
it isentropically to the pressure p u at which it can enter the boiler to 
retraverse the cycle. This compression is accompanied by a slight rise in 
temperature (see Chap. 7) which, for purposes of clarity, has been greatly 
exaggerated in Fig. 14:3. The volume of liquid being small, the (nega¬ 
tive) work of this pump, which is a charge against the work output of the 
prime mover, is very much smaller than that required to drive the vapor 
compressor of the Carnot vapor cycle, and the work ratio of the Rankine 
cycle is correspondingly more favorable than that of the Carnot. This 
makes the efficiency of the cycle less vulnerable to the effects of friction 
and other irreversibilities in the real power plant. The liquid pump is an 


332 


BASIC ENGINEERING THERMODYNAMICS 


adiabatic device, and, again neglecting differences in elevation and veloc¬ 
ity at entrance and exit, the work required to drive it, per pound of liquid 
flow, can be shown to be 

ILE = h x - Ha = hi - h fl ( 14 : 4 ) 


Because the liquid is compressible in only very slight degree, its compressi¬ 
bility is customarily ignored and the liquid-pump process is assumed to 
take place at constant volume. It has been shown previously (see 
Example 3:6 and Art. 9:8) that the work of steady-flow compression 
equals j'v dP per pound of fluid. When compressibility is ignored, v 
becomes a constant and the work of the liquid pump, per pound of fluid, 
may be written as 

w p = v /l f l “dP = (Pu - Pi)vf , (14:5) 

in which v/ t is the specific volume of the saturated liquid at condenser 
pressure. This expression, its value being more easily computed, is 
usually substituted for Eq. (14:4). For the conditions usually existing 
in the real power plant, the error invited is negligible. 

The entire cycle of Fig. 14:3 constitutes a heat-engine cycle, and Eqs. 
(14:1) to (14:5) make it possible to write an expression for its efficiency 
(which is also, by convention, the thermal efficiency of the prime mover) 
based on Eq. (2:14) or on Eq. (2:15), as follows: 


IT (cycle) W e II 


J 


v = 


j ~ Y = (h 2 - h t ) - (hi - h fl ) 

IT _ (h 2 — /is) (hi — hj t ) 

hi hi 

_ h<2 — h‘j — (P u — Pi)v f i/J 

(P u ~ Pi)v fl 


— ho — h 3 — 


JQt 


hi — hfi — 


J 


Qs Qr _ (h% — hi) — (/13 — hfi) 

” Qs /l 2 - /*! 

_ hi — hz — (P u — Pi)VfJJ 

• J 


(Pu - Pl)v fl 

J 


[Eq. (2:14)] 


( 14 : 6 ) 


[Eq. (2:15)] 


It will be noted that the Carnot vapor cycle of Fig. 14:1 also operated 
between limiting pressures p u and p h these being the saturation pressures 
equivalent to the temperatures T s and T R , respectively. This Carnot 
cycle is a reversible cycle when the temperatures of the source and of the 
refrigerator are constant. Under the same conditions, the Rankine cycle 
is not a reversible cycle for heat must flow downhill during the process 















POWER—VAPOR SYSTEMS 


333 


1 -c of Fig. 14:3, i.e., across a finite interval of temperature. It is only by 
introducing the concept of a variable-temperature source, as in Chap. 12, 
that the Rankine cycle may be considered as a reversible cycle so that the 
advantages that go with plotting it on a Ts diagram may be fully realized. 
When this is done, a comparison of the proportionate work and heat- 
supply areas of Figs. 14:1 and 14:3 (areas 1-2-3-4 and 1-2 -b-a of Fig. 14:1 
and areas l-c-2-3-4 and l-c-2-b-a of Fig. 14:3) will show that, for the same 
limiting pressures (and thus limiting temperatures), the cycle efficiency 
of the Rankine cycle will be less than that of the Carnot cycle. Further 
examination shows that the discrepancy between the efficiencies of the 
two cycles results from the slope of the saturated-liquid line; if this line 
were vertical, the Rankine cycle would be identical with the Carnot. A 
vertical liquid line on the Ts diagram is a thermodynamic impossibility, 
for it would mean that the specific heat of the liquid was zero, but a low 
specific heat of the liquid (in proportion to the heat required for vaporiza¬ 
tion of the liquid) would increase the proportion which Rankine-cycle 
efficiency bears to the efficiency of the Carnot under equivalent supply 
and exhaust conditions. Further study of the two figures brings out the 
following additional items: 

1. The discrepancy between the efficiencies of the two cycles will 
increase as p u is increased and/or pi is decreased. This is due to the 
greater proportion which the heat supplied to raise the temperature of the 
entering fluid iy the Rankine-cycle boiler bears to that required to vapor¬ 
ize it at constant temperature. 

2. The efficiency of both cycles will be increased as p u is raised. This 
increase of efficiency will not be directly due to the raising of p u but 
because of the increase in T s which accompanies it. But Ts cannot be 
increased above the critical temperature of the working fluid ( T g of Figs. 
14:1 and 14:3) without entirely eliminating the efficient addition of heat 
at constant temperature. The saturated-liquid and saturated-vapor 
curves of Figs. 14:1 and 14:3 have been located and proportioned on the 
assumption that water is the working fluid. The possibility of using 
some more desirable fluid will be discussed later in this chapter. 

3. The efficiency of both cycles will be increased as pi is lowered. 
Again, this increase of efficiency will be due directly not to the lowering of 
Pi but to the decrease of T R which is associated with it. The available 
refrigerator is the atmosphere, and it is obvious that T R cannot be lowered 
below atmospheric temperature. This sets a lower limit for pi which is 
again based on the pressure-temperature relation of the working fluid. 

4. The proportions of Figs. 14:1 and 14:3, when compared, make it 
evident that the efficiency of the Rankine cycle is, under equivalent sup¬ 
ply and exhaust conditions, a quite respectable fraction of the limiting 
efficiency as set by the Carnot cycle. This proportion of ideal efficiency 


334 


BASIC ENGINEERING THERMODYNAMICS 


is certainly much higher than that for the gas cycles discussed in Chap. 12 
(see Fig. 12:3, for example), and it would appear that the Rankine-cycle 
power plant should operate at much higher efficiency than the internal- 
combustion power plant. As a matter of fact, the highest efficiencies 
obtainable in the operation of real vapor power plants based on the 
Rankine cycle are somewhat below those yielded by the best internal- 
combustion plants. This results from the use of external combustion in 
heating the fluid in the Rankine-cycle plant, causing the metallurgical 
limit to step in to fix the highest temperature that can be attained by the 
fluid. It has been seen that the source temperature that is available 
through the process of combustion is between 3000 and 4000°F, while the 
metallurgical limit for the materials normally used in the construction of 
boilers is about 1000°F. The temperature T s in the Rankine-cycle power 
plant is thus limited in practice to a much lower level than when internal 
combustion is employed. In other words, the irreversibility of the vapor 
plant with respect to the difference between the available temperature of 
the source and the highest temperature reached by the working fluid is 
much the greater, and although the Rankine cycle shows a much higher 
proportion of Carnot-cycle efficiency when compared on the basis of the 
same T s , the comparison is made with a much less efficient Carnot cycle 
than is the case for the internal-combustion plant. On the other hand, a 
wider selection of fuels is available for external combustion, and the cost 
of fuel per Btu of calorific value may accordingly be lower. 

5. In Figs. 14:1 and 14:3, condition 2 at exit from the boiler has been 
shown as a saturated-vapor state. In practice, the vapor usually con¬ 
tains at least a small percentage of the liquid at exit from the boiler so 
that its quality is slightly less than 1 and point 2 would lie slightly to the 
left of the position shown in the figures. The efficiency of the Carnot 
cycle would not be changed by this alteration since the proportion which 
the work area bears to the heat-supply area remains unchanged though 
both areas are reduced. The efficiency of the Rankine cycle would 
decrease slightly because the ratio of heat supplied at constant tempera¬ 
ture to total heat supplied is slightly smaller, i.e., the average temperature 
at which heat is supplied is slightly lower. 

Example 14:24. A Rankine cycle, using steam as the working substance, operates 
between limiting pressures of 600 and 1 psia. The steam enters the prime mover 
as a saturated vapor, (a) What is the net work of the cycle per pound of steam 
flowing through the cycle? ( b ) What is the efficiency of the cycle? (c) What is 
the moisture percentage in the exhaust steam as it leaves the prime mover? 

Solution: 

(a) hi = h fl = 69.7; h 2 = 1203; h z = 807. 


-j = h 2 - hs = 1203 - 807 = 396 Btu/lb 


[Ex. 14:1] 


POWER—VAPOR SYSTEMS 


335 


Note that this is the same as for the Carnot-cycle prime mover of Example 14:1. 
y 4 = Vfl = 0.01614 ft 3 /lb 

W p = (P u - Pi)v fl = 144(600 - 1)(0.01614) = 1390 ft-lb or 1.8 Btu/lb. 

[Eq. (14:5)] 

The net work of the cycle is 


W e W p 


J 


J 


= 396 - 1.8 = 394.2 Btu/lb 


(b) Applying Eq. (14:6) to calculate the cycle efficiency, 


V = 


1203 - 807 - 1.8 394.2 

1203 - 69.7 - 1.8 1131.5 


= 0.348 


(c) The condition of the steam as it leaves the prime mover is indicated by the position 
of point 3, corresponding to a pressure of 1 psia and an enthalpy of 807 Btu. The 
moisture percentage may be conveniently read from the Mollier chart as 28.8 per cent. 

It is sometimes desirable, for practical reasons, to deliver superheated 
vapor to the prime mover. For example, when a turbine is used as the 
prime mover, the position of point 3 to 
the left of the dry-vapor line indicates 
that there is considerable moisture in 
the steam jet as it impinges on the 
blades, and this moisture may be the 
cause of serious blade erosion, as has 
been pointed, out in Art. 11:12. If 
point 2 is moved to the right, into the 
superheat region, point 3 will also 
move to the right and its new position 
will indicate that a lower proportion of 
dense liquid particles is carried in the 
high-velocity steam jet. 

In the Carnot cycle of Fig. 14:1, 
point 2 cannot move to the right with¬ 
out abandoning the practical require¬ 
ment that heat be supplied the cycle 
at constant pressure. On the other hand, in the Rankine cycle it is 
specified that heat shall be supplied at constant pressure and point 2 of 
the Rankine cycle, with superheat, will lie as shown in the Rankine cycle 
of Fig. 14:5. The flow diagram for this cycle is similar to Fig. 14:4 but 
with the addition of a heat exchanger, called the superheater, between 
boiler and prime mover. This is necessary since large amounts of super¬ 
heat cannot be given the vapor while it is in close contact with its liquid. 
The superheater accepts the vapor from the boiler at, or near, state d, and 
process d- 2, a constant-pressure process, is carried out in that unit, the 
vapor being discharged as a superheated vapor at state 2. Equation 



heat. 













336 


BASIC ENGINEERING THERMODYNAMICS 


(14:6) is still valid for the calculation of the cycle efficiency when the 
change in position of points 2 and 3 is taken into consideration. The 
movement of state 3 to the right and thus to lower percentages of the 
liquid in the exhaust vapor will be noted. 

A comparison of the proportions of Figs. 14:3 and 14:5, both Rankine 
cycles having the same limiting pressures p u and pi, shows that the cycle 
of Fig. 14:5 will have the higher efficiency and therefore would seem to be 
the more thermodynamically advantageous. But point 2 cannot lie 
above the temperature extreme set by the metallurgical limit in either 
cycle, and a still higher efficiency could have been realized if the heat had 
been supplied at constant temperature (as to a saturated vapor) at the 
temperature level reached as the result of the superheating process. 
Thus the cycle of Fig. 14:5 compares less favorably with its equivalent 
Carnot cycle than does the cycle of Fig. 14:3. When a fluid with a critical 
temperature which is low as compared with the metallurgical limit, such 
as steam, is used, the use of superheat may be thermodynamically desir¬ 
able. Superheat would not be used with fluids having more desirable 
characteristics than steam in this respect, except for practical reasons 
such as the damage that may be caused in the turbine prime mover by 
excessive moisture percentages in the steam. 

Example 14:22?. The cycle is the same as that of Example 14:24. except that the 
steam is superheated to 800°F before it enters the turbine prime mover. Answer 
the same questions that are asked in that example. 

Solution: 


(a) hi = 69.7 Btu and W v /J = 1.8 Btu as in Example 14:24. From the Mollier 
chart, h 2 = 1408 Btu, and h 3 = 914 Btu. 

W 

-j- = 1408 - 914 = 494 Btu/lb 

Net work of cycle = 494 — 1.8 = 492.8 Btu/lb 


1408 - 914 - 1.8 
1408 - 69.7 -178 


492.2 

1336.5 


0.368 


This efficiency is higher than that of the Rankine cycle of Example 14:24. However, 
that cycle compared with a Carnot cycle having an efficiency of 0.407 (see Example 
14:1), whereas the efficiency of the cycle of this example should logically be compared 
with that of a Carnot cycle to which heat was supplied at 800°F; the corresponding 
Carnot cycle efficiency would be (800 — 101.74)/(800 + 460) = 0.555. 

(c) The moisture percentage in the steam as it leaves the prime mover may be read 
from the Mollier chart at state 3 and is found to be about 18.5 per cent. This is an 
improvement over the corresponding result in Example 14:24 if the prime mover is 
of the turbine type. 


14:3. The Reheat Cycle. The maximum percentage of moisture that 
is considered permissible in the exhaust steam, if excessive blade erosion 
is to be avoided when a turbine is the prime mover, is about 10 per cent. 





POWER-VAPOR SYSTEMS 


337 


The discussion of Art. 14:2 has indicated that it is advantageous to oper¬ 
ate at as high an upper pressure p u of the Rankine cycle as is possible con¬ 
sidering the pressure-temperature relation of the fluid and its critical 
temperature. For water, the critical temperature is 705°F at a critical 
pressure of 3206 psia, and let us suppose, for the purpose of illustration, 
that it is decided to supply steam to the turbine as superheated steam at a 
pressure of 2600 psia and a temperature (fixed by the metallurgical limit) 
of 1000°F. Reference to the Mollier diagram for steam shows that this 
corresponds to an enthalpy h 2 (Fig. 14:5) of 1456 Btu. We shall further 
assume that the exhaust pressure is 1 psia, corresponding to a saturation 
temperature of about 102°F and conforming approximately with the 
usual practice in the large central-station power plant. The Mollier 


/ Re heater 



chart shows that state 3, at the end of an isentropic expansion to this 
exhaust pressure, is identified as having an enthalpy hz of 850 Btu and a 
moisture percentage of nearly 25 per cent. Allowing for an assumed 
turbine efficiency of 85 per cent, this enthalpy would be increased to 
1456 — 0.85(1456 - 850) = 941 Btu, which, at 1 psia, corresponds to a 
moisture proportion of 16 per cent. This is still a higher moisture per¬ 
centage than is considered permissible, and it appears that we must con¬ 
tent ourselves with the lower cycle efficiency that goes with lower steam- 
supply pressure. After a number of trial calculations, cariied out in a 
manner similar to that used above, we find that if p u — 1200 psia and 
steam is supplied the turbine at that pressure and superheated to the 
metallurgical limit of 1000°F, h 2 = 1499 Btu, h* (at the end of isentropic 
expansion to 1 psia) =911 Btu, and, again assuming a turbine efficiency 
of 85 per cent, the enthalpy of the exhaust steam is 1499 - 0.85 (1499 
_ 911) = 999 Btu, corresponding to a moisture percentage of about 10 



























338 


BASIC ENGINEERING THERMODYNAMICS 


Meta/Iur gica/ firrtif’ 


per cent. It thus appears that a limit is set on the attainment of higher 
efficiency through raising the upper pressure of the Rankine cycle by the 
necessity of avoiding excessive blade erosion when the turbine is used as 
the prime mover. Since the turbine is the accepted type of prime mover 
for the large central-station plant, this is a serious limitation; the reheat 
cycle is used to avoid it. 

The flow diagram of the reheat cycle is shown in Fig. 14:6. The boiler 
and superheater operate at the upper pressure p u of the cycle, which may 
exceed the limit of 1200 psia calculated above. The first stage of expan¬ 
sion is carried out in the high-pressure turbine, which expands the steam 

to some intermediate pressure pi at 
which the moisture that has accumu¬ 
lated has not exceeded the allowable 
limit. The steam exhausted from this 
unit then enters the reheater (essen¬ 
tially a superheater in its design fea¬ 
tures) and is brought up to as high a 
temperature as is practical before fur¬ 
ther expansion in the low-pressure tur¬ 
bine to final exhaust pressure pi. The 
ideal reheat cycle is shown on Ts coor¬ 
dinates in Fig. 14:7. If the initial 
expansion had been carried through, 
without reheating, the final quality of 
the steam would have been as at point 
/ of this diagram. Because of reheat, 
the state of the exhaust steam shifts to 
point 5, corresponding to a lower per¬ 
centage of moisture. 

A comparison of Figs. 14:6 and 14:7 shows that, per pound of fluid flow, 
the heat supplied in the boiler and superheater is h 2 — hi = h 2 — h fl 

- — j—Qs will also include the additional heat supplied in the 



Fig. 14:7. The reheat cycle. 


process carried out in the reheater, which amounts to h A — hz. The heat 
Q r rejected through the action of the condenser, which carries out process 
5-6, is h$ — h 6 = h*> — hf r The work of the high-pressure turbine is 
h 2 — hz, and that of the low-pressure turbine is h± — h b . The work that 
must be supplied to drive the feedwater pump is ( P u — Pi)v fl . Based on 
Eq. (2:14), 


W 

J Qs 


(h 2 — hz) + (hi — hz) — -—-— -j — 

(h 2 - h„ - + ( h * - h,) 


(14:7) 






















POWER—VAPOR SYSTEMS 


339 


The efficiency of the reheat cycle will be found to agree quite closely with 
the efficiency of the simple Rankine cycle formed by continuing the origi¬ 
nal expansion to the final exhaust pressure (cycle l-c-d-2-/-6 of Fig. 14:7), 
but it avoids the high moisture percentages in the exhaust steam of that 
cycle. While the use of reheat does not in itself materially increase the 
efficiency of the cycle, it enables the use of higher steam-supply pressures 
and thus the employment of a more efficient cycle. 

Example 14:3. Steam is supplied the high-pressure turbine of a reheat-cycle plant 
at 2600 psia, 900°F. After expansion in that unit to a pressure of 400 psia, the steam 
is reheated to 900°F and expands in the low-pressure turbine to a final pressure of 
1 psia. (a) What is the efficiency of the cycle? Compare this efficiency with that 
of the Rankine cycle which is produced by continuing the original expansion to the 
condenser pressure of 1 psia (to point /). (6) What are the moisture percentages in 

the steam as it leaves the final stages of the turbine prime mover in each case? (c) 
Allowing for an efficiency of the low-pressure turbine of 85 per cent, what is the mois¬ 
ture percentage at this point in the reheat cycle? 

Solution: 

(a) The notation is that of Fig. 14:7. From the Mollier diagram, h 2 = 1384 Btu; 
h 3 = 1192 Btu; hf = 821 Btu; hi = 1470 Btu; h 3 = 964 Btu. 

z 3 = 0.983: x f = 0.725; x 5 = 0.863 
From the tables, h 6 = h/ t = 69.7 Btu and = Vf t = 0.01614 ft 3 /lb. 

Pump work *= W p = (P u - Pi)v fl = 144(2600 - 1)(0.01614) = 5820 ft-lb, or 

7.5 Btu /lb 


Applying Eq. (14:7) to calculate the efficiency of the reheat cycle, 

(1384 - 1192) + (1470 - 964) - 7.5 690.5 

v ~ (1384 - 69.7 - 7.5) + (1470 - 1192) 1584.8 

The efficiency of the Rankine cycle, without reheat, is, from Eq. (14:6), 

1384 - 821 - 7.5 _ 555.5 _ 
v ~ 1384 - 69.7 - 7.5 1306.8 

It will be noted that the difference between the efficiencies of these two cycles is not 

large. 

(6) As the result of isentropic expansion, the moisture percentages read from the 
Mollier chart are 13.7 and 27.5 per cent for the reheat cycle and the Rankine cycle, 
respectively. 

(c) The enthalpy at the end of expansion in a real low-pressure turbine which has 
a turbine efficiency of 85 per cent will be designated as h b >. Then h b > = 1470 — 
0.85(1470 — 964) = 1040 Btu, and this enthalpy corresponds, at a pressure of 1 psia, 
to a moisture percentage of 6.4 per cent. This is less than the permissible limit of 
10 per cent. 

14:4. The Regenerative Vapor Cycle. The Rankine cycle, without 
superheat, has an efficiency less than that of the equivalent Carnot solely 
because of the lower average temperature at which heat is received from 






340 


BASIC ENGINEERING THERMODYNAMICS 


the source during the process of heating the liquid to saturation tempera¬ 
ture. This is somewhat the same problem that was encountered in the 
Stirling cycle discussed in Art. 4:10, and it will be remembered that the 
efficiency of that cycle was raised to equal that of the equivalent Carnot 
by employing the principle of regeneration, i.e., by removing heat from 
the working fluid during one of the processes of the cycle and utilizing this 
heat to accomplish the rise in temperature required during a later process 
of the cycle; thus the heat from the source may be supplied at constant 
temperature, and the cycle may attain the maximum conceivable effi¬ 
ciency, that associated with the Carnot 
cycle. 

The regenerative vapor cycle applies 
regeneration to the liquid-heating proc¬ 
ess (1-c of Fig. 14:3). The heat nec¬ 
essary to accomplish this rise in tem¬ 
perature of the liquid is removed from 
the vapor during its expansion through 
the prime mover. The resulting cycle', 
as it appears when a saturated vapor 
is supplied the prime mover, 1 is pre¬ 
sented in Fig. 14:8. In this figure, the 
crosshatched area 2-k-h-e represents the 
heat removed from the expanding vapor 
and used to supply the heat, repre¬ 
sented by area 1 -c-h-a, necessary to 
preheat the liquid to vaporization temperature; these areas will later be 
shown to be equal and to have the same shape. 2 

1 The regenerative vapor cycle, for reasons to be advanced later, is best applied to 
plants in which the turbine is the prime mover. As has been brought out above, 
superheated vapor is usually supplied the turbine prime mover for practical reasons, 
and this introduces a second irreversibility due to the addition of heat to the working 
fluid at variable temperature during the superheating process. In order to concen¬ 
trate attention on the regenerative heating of the liquid, the cycle here discussed does 
not employ superheat. However, the regenerative-liquid-heating principle may be 
applied to the cycle even when the vapor is supplied the turbine as a superheated 
vapor, although, in that case, a cycle efficiency equal to that of the equivalent Carnot 
cycle cannot, even in theory, be realized. 

2 To prove that line 2-k parallels line 1-c, it is necessary only to show that the slopes 
of these lines are everywhere the same at the same temperature. In the ideal regenera¬ 
tion that is here supposed, the differential of temperature between the vapor which 
yields the heat and the liquid which receives it is infinitesimal. If this infinitesimal 
differential of temperature is to be maintained throughout the length of the counter¬ 
flow regenerative heat exchanger, the transfer of a small quantity of hea dQ from vapor 
to liquid must be accompanied by a decrease of vapor temperature (dT)v which is 
equal to the increase of liquid temperature ( cIT)l, or ( dT)v = — ( dT)i,- Let us 
examine the effect on the entropy of the vapor and of the liquid of the transfer 



Fig. 14:8. Regenerative vapor cycle 
(without superheat). 






POWER—VAPOR SYSTEMS 


341 


Ihe heat Qs received from the source is thus reduced from that required in 
the Rankine cycle of Fig. 14:3 to area c-2-e-h , and Q R , discharged to the 
refrigerator, is represented by area 4-3 -b-a. Because the crosshatched 
areas are the same shape, the widths of the Qs and Q R areas are the same 
and their magnitudes are proportional to the temperatures T s and T R , as 
in the Carnot cycle; thus the efficiency of the regenerative cycle l-c-2-/c-3-4 
is that of the Carnot cycle which operates between the temperatures T s 
and T r . 

The idealized flow diagram of a vapor power plant to operate on the 
cycle of Fig. 14:8 is shown in Fig. 14:9. The liquid pump removes 
saturated liquid at pi and T R from the condenser and raises its pressure 



to p u . The temperature is also raised slightly owing to the compression 
of the liquid, and the liquid goes through process 4-1 of Fig. 14:8. After 
leaving the pump, the liquid passes through the turbine in a direction 
opposite to that of the flow of vapor through that unit. Thus a counter¬ 
flow heat exchange is made possible, and the liquid issues at the high- 
pressure end of the turbine at the temperature T s ', this is process 1-c of 
Fig. 14:8. The liquid then enters the boiler as a saturated liquid and 
evaporates at constant temperature (process c- 2) in that unit as it receives 
heat, in this case from the source. The expansion of the vapor in the 
turbine is no longer isentropic, as it is giving up heat to the liquid for the 
purposes of regeneration, and the expansion line 2 -k moves to the left, 


of this element of heat. For the liquid, heated at constant pressure, we may write 

(dT)L’ f rom the sl 0 p e of the liquid 


( 3Q)l = c pl (cIT)l and dsL = = c 


rji w PL rji 

line 1-c, representing the liquid-heating process on the temperature-entropy diagram, 

= — • For the vapor, ( dQ)v = — (dQ)z, = —c PL (dT)L = c pL (dT)v. Then 
\as J l c pL 

, and the slope of the vapor line 2 -k on the Ts diagram is 


is 


dsv = 


(dQ)v c pL (dT)y 
T 


(dT\ = T = /(iT\ 
\ds Jv c pL \dsjL 
shown, the heat-exchange 


Note that if lines 1-c and k-2 are parallel, as has here been 
areas must have the same shape and si?e, t | 















342 


BASIC ENGINEERING THERMODYNAMICS 


paralleling line 1-c. * The very last stage of expansion in the turbine 
(k to 3) is isentropic, for it will be remembered that the liquid pump has 
raised the temperature of the liquid slightly above T R before it entered 
the turbine for counterflow regeneration. 

The reader may form a mental picture of the regenerative heat ex¬ 
changer of Fig. 14:9 as consisting of liquid-carrying tubes which are 
embedded in the turbine casing and in the walls of the turbine nozzles. 
This is an entirely impractical arrangement; instead, in the real steam 
power plant that operates with regenerative feedwater heating, the heat¬ 
ing of the feedwater is accomplished by steps or stages. In Chap. 11, it 
has been shown that, if desired, some of the steam may be withdrawn 
from the turbine at the end of any turbine stage. In regenerative feed- 
water heating as applied in practice, the extracted steam is conducted to a 
heat exchanger, where it is brought into contact with the feedwater and, 
as it condenses, raises the temperature of the water as the latter flows 
toward the boiler. The pressure of the steam remains constant during 
condensation and therefore, if the steam is saturated as it enters the 
heater, the condensation process is represented by a horizontal line on the 
temperature-entropy diagram. If a large number of extraction points (a 
separate heater is required for each) are selected, line 2 -k becomes a series 
of small steps as indicated in Fig. 14:8. The horizontal section of a step 
represents the condensation of steam in a feedwater heater, the vertical 
section the isentropic expansion of the steam in the turbine between 
bleeding points. As the number of extraction points is increased, the 
stepped expansion line approaches the smooth line 2-k in contour. In 
practice, the number of extraction points (and therefore of heaters) may 
be one to five or six, depending upon the complexity of design which is 
justified. Although the cycle efficiency increases with the number of 
heaters, the gain in efficiency that is produced by the addition of another 
extraction point becomes less and less with each addition; in the mean¬ 
time, practical factors, such as equipment cost, fixed charges, and main¬ 
tenance expense, are increasing and ultimately have the effect of canceling 
the theoretical gains. 

The flow diagram for a three-heater installation is shown in Fig. 14:10; 
again, for simplicity, steam is furnished the turbine of the plant as satu¬ 
rated steam. The extraction points in the flow of steam through the 
turbine are m, n, and o and, in practice, must fall between turbine stages; 
their best location from a thermodynamic viewpoint will be discussed 
later. From each extraction point, the steam is led to a corresponding 
heater, in which, as it condenses, it gives up heat to the feedwater. The 
unextracted portion of the steam continues to expand through the lower 
pressure stages of the turbine and finally reaches the condenser, to be 

*For footnote see pages 340 and 341. 


POWER—VAPOR SYSTEMS 


343 


there condensed into a saturated liquid. This liquid, in the flow arrange¬ 
ment shown in the figure (other pump arrangements are possible and 
constitute slight variations from this cycle), is pumped from the condenser 
and raised to boiler pressure p u by the feedwater pump. Instead of going 
directly to the boiler, the feedwater passes through the heaters H 0 , H n , 
and H m in that order (the reverse of the order in which steam has been 
extracted from the turbine). An increment of heat is added to the feed- 
water in each heater, bringing its temperature, in the limit, up to the 
temperature of the condensing steam in that heater and thus to the satu¬ 
ration temperature equivalent to the pressure at the respective extraction 
point. The amount of steam extracted to each heater is automatically 



Fig. 14:10. Flow diagram—three-heater regenerative vapor cycle. 


controlled by the condensation within that heater, since any condensation 
will lower the pressure within the heater and thus furnish the pressure 
differential necessary to draw in an additional supply of steam. It is 
therefore adjusted according to the heat given up to the feedwater as the 
latter passes through the heater. A trap is placed below each heater, 
permitting only condensate to leave the heater; this condensate is picked 
up by a pump and raised to supply pressure p u to join the feedwater at a 
point beyond the heater. In the limit, the temperatures of condensate 
and feedwater are the same at this junction, and no irreversibility is 
caused by the mixing of the two streams. 

The temperature-entropy diagram for this three-heater cycle is shown 
in Fig. 14:11. The saturated-liquid and dry-vapor lines have been shown 
on this diagram for 1 lb of steam entering the turbine. The entire pound 
of steam expands isentropically to the first extraction point m. The 
process m-m' of the figure represents the constant-pressure condensation 
of M m lb of extracted steam, which takes place in heater H m . The loca¬ 
tion of m' is fixed by the sum of the entropies of the condensate resulting 
from the condensation of the extracted steam and that of the portion of 














344 


BASIC ENGINEERING THERMODYNAMICS 


the original pound which continues its flow through the turbine; their 
combined weights are 1 lb. The balance of the steam, weighing 1 — M m 
lb per pound of steam entering the turbine, continues its isentropic expan¬ 
sion to the second extraction point n, where a weight M n is extracted to 
heater H n and is condensed therein, accounting for process n-n'; the 
same procedure is followed at each extraction point. 

Returning to heater H m , the feedwater enters this heater at, in the 
limit, a temperature corresponding to the saturation temperature at the 
second extraction point, having been raised to that temperature in heater 
H n . It is raised, in heater H m , to T m , and this feedwater-heating process 

is represented in Fig. 14:11 by line 
n"-m". The heat required for this 
process is represented by 2 Lresin"-m"-r-q 
of the figure and is equal to the heat 
given up through the condensation of 
extracted steam in heater H m , or area 
m'-m-t-s. These two areas therefore 
represent the regenerative heat ex¬ 
change as far as heater H m is concerned. 
Similar paired area strips may be located 
for each heater in the series. These 
regenerative transfers of heat are ir¬ 
reversible because of the increase of 
temperature of the feedwater, but the 
degree of their irreversibility is less 
than if the heat had been supplied by 
the source at the higher temperature Ta. The efficiency of the cycle will 
therefore lie between those of the equivalent Rankine and Carnot cycles. 

In order to calculate the efficiency of the cycle of Figs. 14:10 and 14:11, 
it is necessary to know the respective weights extracted at points m, n, 
and o. It has been explained above that these weights are automatically 
adjusted to the requirements of feedwater heating or, in other words, that 
the heat yielded by the condensing vapor in the heater equals the heat 
absorbed by the feedwater. By an application of Eq. (3:5), it may be 
shown that the heat given up by the vapor is equal to its decrease of 
enthalpy and that the heat absorbed by the liquid is identical with its 
increase in the same property. Using heater H m as an example, the 
weight of steam condensed is M m , its specific enthalpy entering the heater 
being h m , the enthalpy at the end of an isentropic expansion from point 2 
to the pressure p m at the extraction point; this can be conveniently deter¬ 
mined by graphical projection on the Mollier diagram. The specific 
enthalpy of the extracted steam at exit from the heater, as a saturated 
liquid at pressure p m , may be denoted as hf m and can be read from the 



Fig. 14:11. Three-heater 
ative vapor cycle. 










POWER—VAPOR SYSTEMS 


345 


tables. The weight of feedwater passing through the heater is 1 — M m 

(see Fig. 14:10), and its specific enthalpy at entrance is h fn + — J >n ) v f n , 

in which the second term represents the effect of the feedwater pump on 
the enthalpy, in raising the pressure above the saturation pressure. 
Similarly, the specific enthalpy of the feedwater at exit from the heater is 
(JPu — P m )Vf 

h/ m +-- j ——-• Equating the heat given up by the extracted steam 

to that received by the feedwater, we may write 

(Pu - Pm)v fm 


Mm(hm — hf m ) — (1 — Mm) I kf m + 


J 


h fn ~ 


(Pu ~ Pn)v fn ' 


J 


(14:8) 


In this equation, once the extraction points have been selected, all terms 
may be independently evaluated with the exception of the weight of 
steam extracted; thus M m may be computed. In practice, the terms 
representing the effect of feedwater-pump work on the enthalpy are rela¬ 
tively small and, since they add considerably to the complications of solu¬ 
tion without materially changing the final result, are usually dropped. 
Thus Eq. (14:8) is customarily written in the form 


or 


M m (h m hf m ) (1 Mm) (hf m hf n ) 


M m = 


hyf m _ kf n 

hm hjf n 


(14:9) 


A similar analysis to obtain the weight extracted to the second heater is as 
follows: 


M n (h n — hf n ) — (1 — M m — M n )(hf„ — hf 0 ) 


or 


Tir _ (1 M m ) (hf n hf 0 ) 

n L 

h n — hf 0 


(14:10) 


By substituting the value of M m , as expressed in Eq. (14:9), in this equa¬ 
tion, it may be changed to the form 


M n 


h m hf m hfn _ hf 0 

h m h/ n h n hf 0 


(14:11) 


Proceeding in a similar manner, it may be shown that 

nf (1 M m M n ) (hf 0 hf 3 ) _ h m hf m h n hfn hf 0 hf 3 .-.q\ 

M ° “ ho- hf 3 - hm- hf n hn - h fo h 0 -hf 3 K J 

In substitution in Eqs. (14:9) to (14:12), it should be carefully noted that 
h m , h n , h 0 , and h z are the specific enthalpies of the steam at the respective 














346 


BASIC ENGINEERING THERMODYNAMICS 


points in the turbine and not at the position of these points as shown in Fig. 
14:11; on that figure they would lie vertically below point 2. They are 
conveniently obtained from the Mollier diagram by continuing the same 
vertical projection from state 2 that was used to determine h m to intersect 
the pressures at successively later extraction points and the pressure at 
the turbine exhaust. The extraction of steam for feedwater heating does 
not reduce the quality of the steam flowing through the turbine to the con¬ 
denser, but merely its quantity. 

Having made these preliminary calculations, the efficiency of this three- 
heater cycle can be obtained by applying Eq. (2:15). The heat Q s sup¬ 
plied by the source heats the liquid from m" to c and then evaporates it to 
state 2; the amount of this heat per pound of steam entering the turbine 


is 


ho — h m " or ho — h. 


(Pu ~ Pm)Vf n 
J 


The amount of heat Q 


R 


rejected to the refrigerator is (1 — M r 
cycle efficiency is 


M n — M 0 )(h d — h f9 ), and the 


v = 


Qs — Q 


R 


Qs 
h 2 - h f „ 


(P u - Pm)v f „ 

J 


- (1 - M m - M n - Mo)(h 3 - h /M ) 


ho — hf m 


0 u P m)Vf„ 

~fT 


(14:13) 


Although the calculations outlined in Eqs. (14:8) to (14:13) have here, for 
clarity, referred to the three-heater cycle of Figs. 14:10 and 14:11, the 
same general method of calculation will apply to a cycle with any number 
of heaters. The pump arrangement shown in Fig. 14:10 is only one of a 
number of plans that can be used; for example, by draining the condensate 
from each heater into the heater at next lower pressure and, eventually, 
into the condenser, the number of pumps can be reduced to one, and the 
weight of condensate passing through each heater is the same. Note that 
this change would lower the efficiency of the cycle slightly, since the mix¬ 
ing of the condensate with steam at lower temperature will introduce an 
additional irreversibility which is avoided in the cycle as discussed above. 

The irreversibility in the cycle which has been used here has arisen from 
the difference between the temperature of the steam in the heaters and the 
temperature of the feedwater as it entered the heater; otherwise the cycle 
efficiency would have equalled that of the Carnot cycle. The selection 
of the best points at which to extract steam from the turbine is based on 
keeping the average differential of temperature in the heaters to a mini¬ 
mum. This can best be approximated by dividing the total temperature 
interval between the saturation temperatures corresponding to the supply 
pressure p u and exhaust pressure pi into equal temperature drops between 










POWER—VAPOR SYSTEMS 


347 


inlet and the first extraction point, between extraction points, and 
between the last extraction point and the turbine exhaust. Thus it will 
be observed that, in Fig. 14:11, the length of the verticals 2 -m, m'-n, n'-o , 
and o -3 have been made equal. Note that the number of these successive 
drops is one greater than the number of heaters. Another factor to be 
considered before a final selection is made is the design of the turbine, 
since an extraction point may be located only between stages of that unit. 

When water is the working fluid and the temperature of the exhaust 
steam is not far above atmospheric temperature, a secondary advantage is 
associated with the regenerative vapor cycle. For example, at 100°F the 
saturation pressure of water is less than 1 psia, and the specific volume of 
dry saturated steam is about 350 ft 3 /lb. This compares with a specific 
volume of saturated steam at a supply pressure of 600 psia, corresponding 
to a saturation temperature of about 486°F, of about 0.77 ft 3 /lb, and indi¬ 
cates what a tremendous increase of volume has taken place during expan¬ 
sion through the turbine, even when allowance is made for the lower qual¬ 
ity of the steam at exhaust. The very great volumes of steam that must 
be handled near the exhaust end of the turbine add to the difficulties of 
design of its low-pressure stages and to the size and cost of the turbine. 
In the regenerative cycle, the weight of steam flow is less through the 
lower pressure stages; this may amount to relieving these stages of one- 
quarter or more of the steam flow that must be handled at entrance to the 
turbine, and the difficulty mentioned above finds a partial solution. 

The reheat and the regenerative cycles are both typically applied when 
the turbine is the prime mover and when the difference between Ts and 
T r is large. The advantages of both cycles may be secured by combining 
their essential features in a single cycle, called the reheat-regenerative 
cycle. The supply steam enters the high-pressure turbine of Fig. 14:6 
and may pass one or more extraction points before the unextracted portion 
is removed and reheated. Additional extraction points are usually 
located in the low-pressure turbine. 

Example 14:4. A three-heater regenerative vapor cycle supplies saturated steam to 
the turbine at 600 psia. The exhaust pressure is 1 psia. (a) Choose suitable extrac¬ 
tion points, and determine the specific enthalpy of the steam at each of these points 
and at entrance to and exit from the turbine. (6) Neglecting the effect of pump work, 
find the specific enthalpy of the feedwater at exit from the feedwater pump and from 
each of the regenerative feedwater heaters, (c) Calculate the weight of steam 
extracted to each heater per pound of steam entering the turbine, (d) Calculate the 
efficiency of the cycle, and compare it with those of the equivalent Carnot and Rankine 
cycles of Examples 14:1 and 14:2A. 

Solution: 

(a) The saturation temperatures equivalent to steam pressures of 600 and 1 psia are, 
respectively, 486.21 and 101.74°F, the total temperature interval being approximately 
382 deg. Dividing this interval by one more than the number of heaters, —= 95.5°. 


348 


BASIC ENGINEERING THERMODYNAMICS 


Subtracting 95.5 from 486.2, the temperature at the first extraction point should be 
about 391°F, corresponding to a saturation pressure of around 225 psia. This pres¬ 
sure will be selected, in the absence of information as to the design of the turbine, as 
the pressure at the first point at which steam is extracted from the turbine. Con¬ 
tinuing the same procedure, the temperatures at the second and third extraction 
points are found to be about 295 and 200°F, corresponding to pressures of about 62 and 
12 psia. The specific enthalpies are obtained by vertical projection on the Mollier 
chart from state 2 and are as follows: h 2 = 1203; h m = 1123; h n = 1030; h 0 = 930; 
hz = 807. 

( b ) The enthalpies of the feedwater at exit from the pump and from each heater are: 


At exit from pump (hf at 1 psia) = 69.7 = h/ 3 

At exit from heater H 0 (hf at 12 psia) = 170.0 = hf 0 

At exit from heater H n (hf at 62 psia) = 264.3 = hf n 

At exit from heater H m (hf at 225 psia) = 366.1 = h/ m 


( c ) From Eqs. (14:9), (14:11), and (14:12), 


Mm 

M n 

Mo 


366.1 - 264.3 101.8 


1123 - 264.3 
1123 - 
,1123 
/1123 - 366.1\ / 
\1123 - 264.3 / \ 


/1123 - 366.1\ / 
\1123 - 264.3/ \ 


858.7 

264.3 


= 0.119 lb 
- 170.0 


= (0.881)(0.110' 

- 264.3X /170.0 - 69.7\ 

- 170.0/ V 930 - 69.7 ) 


1030 - 170 
1030 
1030 


(d) Applying Eq. (14:13) to compute the cycle efficiency, 


= 0.097 lb 

(0.881) (0.891) (0.1163) 
= 0.091 lb 


(Pu - Pm)v fm 144(600 - 225) (0.01852) 
J 778 


1.3 Btu 


and 

1203 - 366.1 - 1.3 - (1 - 0.119 - 0.097 - 0.091) (807 - 69.7) 324 

v ~ 1203 - 366.1 - 1.3 ~ 835.6 

= 0.388 

This represents a considerable step from the Rankine-cycle efficiency of 0.348 (Exam¬ 
ple 14:2A) toward the limiting Carnot-cycle efficiency of 0.407 (Example 14:1). 

Checking, the net work of the cycle is Qs — Qr, or 324 Btu/ lb, from the calcula¬ 
tion for efficiency above. The gross work per pound of steam entering the turbine 
may be calculated as 


y = 1(1203 - 1123) + (1 - M«)(1123 - 1030) + (1 - M m - M n )(1030 - 930) 

+ (1 - Mm - M n - Mo) (930 - 807) 
= 80 + (1 - 0.119)(93) + (1 - 0.119 - 0.097)(100) 

+ (1 - 0.119 - 0.097 - 0.091) (123) 

= 80 + 82 + 78.4 + 85.3 = 325.7 Btu 


From this gross work must be subtracted the sum of the work of the pumps. The 
work of the feedwater pump is 


(1 - Mm - M n - Mo)(Pu - Pi)v fl _ (0.693) (144) (600 - 1) (0.01614) 
J 778 


1.2 Btu 


The work of condensate pump P m below heater H m is 

Mm(Pu - Pm)Vf m (0.119) (144) (600 - 225) (0.01852) ^ _ 

J = - 178 - = °' 2 Btu 

















POWER-VAPOR SYSTEMS 


349 


The work of pump P n is 

Mn(Pu P n)V f n 

J 

The work of pump P c is 

Mo(Pu — Po)Vfo 

J 

The sum of the pump work is 1.8 Btu per pound of steam entering the turbine, and the 
net work of the cycle, calculated by subtracting pump work from the gross output of 
the turbine, is 325./ 1.8 = 323.9 Btu. This checks closely the value of 324 Btu 

obtained as the difference between Q s and Q R . 

14.5. The Ideal Vapor for the Heat Engine. Water is the working 
fluid in the overwhelming majority of real heat engines that employ a 
vapor. But we have had occasion, in preceding pages, to point out a 
number of respects in which the characteristics of water are not those 
which we should prefer in the working fluid, and it is possible that some 
other vapor may be found which will better meet the requirements of a 
fluid for power generation. In order to systematize our search for this 
ideal vapor, let us set down the characteristics which it should possess. 

1. It must be stable under all conditions through which it is to pass so 
that it will not dissociate. 

2. It must not react chemically with the materials which, during 
various processes of the cycle, are called on to contain it. 

3. It must not dissolve these materials with which it comes in contact 
in the course of the cycle. 

4. It should be abundant in nature, cheap, and nonpoisonous. 

5. Its critical temperature should exceed the metallurgical limit by a 
comfortable margin. 

6. Its saturation pressure at the metallurgical limit should be moder¬ 
ate. Thus extremely high stresses in the walls of containers such as the 
boiler and superheater are avoided. 

7. Its triple-point temperature should be below the temperature of 
the atmosphere (the temperature of the available refrigerator). Other¬ 
wise, condensation at atmospheric temperature might produce a solid, 
which would be difficult to transport through succeeding cycle processes. 

8. Its vapor pressure at atmospheric temperature should not be too 
low. If below that of the atmosphere, noncondensable gases from the 
atmosphere may leak into the exhaust line between prime mover and 
condenser, and these must be pumped out of the condenser at considerable 
expense. Also, very low pressures mean very high specific volumes; it has 
been pointed out in the preceding article how large volumes add to the 
size and cost of equipment. 

9. Its change of enthalpy during vaporization should be large as com¬ 
pared with the specific heat of the liquid. This causes the ftankine cycle 


(0.097) (144) (600 - 62) (0.0174) 

778 


= 0.2 Btu 


(0.091) (144) (600 - 12) (0.01665) _ 02 Btu 
778 1.8 Btu 







350 


BASIC ENGINEERING THERMODYNAMICS 


to conform more closely to the Carnot in its efficiency and makes unnec¬ 
essary the expensive equipment required for regenerative feedwater 
heating. 

10. Its entropy, as a saturated vapor at refrigerator temperature, 
should be only slightly greater than as a saturated vapor at the tempera¬ 
ture at which it enters the prime mover. Taking into consideration the 
inevitable irreversibilities that accompany expansion in the prime mover, 
this would prevent either the accumulation of excessive liquid percentages 
in the exhaust or the rejection of some heat at a temperature above the 
available sink temperature which would accompany the constant-pressure 
cooling and condensation of superheated exhaust vapor. The effect of 
this characteristic would be to make it possible to supply the vapor to the 
turbine prime mover as a saturated, instead of a superheated, vapor and 
thus to take full advantage of the metallurgical limit; it would make 
unnecessary the expense of reheat-cycle equipment. 

No one vapor satisfies all of the requirements listed above. Water 
fulfills specifications 1, 2, 3, 4, 7, and 9. With respect to items 8 and 10, 
its characteristics are fairly suitable, but it entirely fails to satisfy 5 and 6. 
Although a vapor which will improve on the characteristics of water over 
the entire range of temperatures between the metallurgical limit and 
atmospheric levels is not available, certain fluids have been suggested 
which have a more desirable behavior over more limited ranges of tem¬ 
perature and the possibility of using these in combination with water in a 
multivapor cycle has received considerable attention. 

14:6. The Mercury-Steam Binary-vapor Cycle. Undoubtedly the 
most important of the fluids that have been suggested for use as a vapor 
for power generation, from the standpoint of the effort devoted to develop-' 
ing its use in a binary-vapor cycle with water as the other fluid, is mercury. 
Mercury has a critical temperature in excess of 2800°F, a triple-point tem¬ 
perature of — 38°F, and, at 1000°F, its saturation pressure is about 180 
psia. It satisfies specifications 1, 2, 3, 5, 6, and 7 of Art. 14:5 quite 
adequately, and its characteristics are fairly satisfactory with respect to 
9 and 10. It is not abundant in nature or cheap and, as a vapor, can be 
highly poisonous. Moreover, it entirely fails to meet the requirements of 
item 8, and this restricts its practical use to only the higher range of tem¬ 
perature since, at condenser temperatures approaching that of the atmos¬ 
phere, it would be entirely impossible to hold the necessary condenser 
vacuum or to handle the tremendous volumes of mercury vapor that 
would be generated. 

A number of power plants based on the use of mercury over the higher 
range of temperatures with steam as the fluid over the lower range have 
been developed and built by the General Electric Company. Some of 
these are now in semiexperimental operation, but equipment of this type 


POWER—VAPOR SYSTEMS 


351 


is not at present sold commercially because of considerations based on 
safety, dependability, and cost. The efficiencies realized have been 
encouraging, being as much as 20 per cent above those obtained in the 
operation of comparable plants using steam alone as the vapor. 

A flow diagram for the mercury-steam binary-vapor power plant in its 
elementary form is shown in Fig. 14:12. Liquid mercury enters the 
mercury boiler and is evaporated to a saturated vapor. This evaporation 
may take place, as a limit, at the temperature fixed by the metallurgical 
limit of 1000°F, corresponding to a mercury-vapor pressure of 180 psia. 
After expansion to a suitable pressure of, say, 2 psia, corresponding to a 





Fig. 14:12. Flow diagram—mercury-steam binary-vapor cycle. 


temperature of about 505°F, in the mercury turbine, the exhaust mercury 
flows to a mercury condenser and is condensed. This mercury condenser 
is also a steam boiler, and, assuming a reasonable temperature differential 
between the mercury and steam to effect heat exchange between the two 
at the desired rate, steam evaporation could take place at a temperature 
of 486°F, corresponding to a steam pressure of about 600 psia. The 
mercury liquid, after leaving the condenser, passes through a pump which 
raises its pressure and forces it into the boiler, completing the cycle of the 
mercury. 

The steam, leaving the steam boiler-mercury condenser unit, passes in 
turn through a steam turbine and steam condenser and is returned by a 
pump as feedwater to that unit; this constitutes a cycle of the water and 
steam entirely similar to cycles discussed earlier in this chapter, except as 
to the source of heat for evaporation in the boiler. If the steam is to 
leave the mercury condenser as a saturated vapor, the rates of flow of 
mercury and steam through their respective cycles must be proportioned 
so that the heat required to evaporate the feedwater into a saturated 






















352 


BASIC ENGINEERING THERMODYNAMICS 


vapor will'be available from the heat given up by the mercury in con¬ 
densing. Since mercury has much smaller enthalpies of vaporization 
than water, this requires that the flow of mercury be at a rate several 
times that of the flow of steam. 

For purposes of clarity and an uncomplicated discussion, the binary- 
vapor power plant of Fig. 14:12 has been shown in its simplest form. 
Improvements in the efficiency of the plant and in practical features, such 
as the moisture percentage in the exhaust steam, could, for example, be 



Fig. 14:13. Mercury-steam binary-vapor cycle. 


made by allowing the saturated steam, after leaving the mercury con¬ 
denser-steam boiler, to pick up further heat by bringing it into communi¬ 
cation with the hot gases, produced as a result of the combustion of the 
fuel, as these gases leave the mercury boiler; or regenerative feedwater 
heating may be incorporated into the steam cycle, with a resultant 
improvement in the efficiency of that cycle and thus of the plant as a 
whole. 

In Figure 14:13, the two cycles described above have been shown on the 
same temperature-entropy diagram. The cycle of the mercury is 
abode , and that of the steam is fghij. The mercury-evaporation process 
cd is shown as extending between the saturated-mercury-liquid and 
saturated-mercury-vapor lines, but these lines are located not as for unit 
weight of mercury but with respect to a weight sufficient to cause the heat 














POWER—VAPOR SYSTEMS 


353 


rejected by the mercury in the mercury condenser-steam boiler (repre¬ 
sented by area aenl ) to be equal to the heat absorbed by the unit weight of 
water and steam passing through the same unit (this heat is shown in the 
figure as area ghijmk). thus it is total entropies rather than the specific 
values of this property that are plotted in locating points a to e. The 
weight of mercury flow per pound of steam flow may be found by com¬ 
paring h e — h a (for 1 lb of mercury) with hi — h g (for 1 lb of steam). 
This ratio of mercury flow to steam flow will be affected by introducing 
the practical factor of mercury-turbine efficiency and the effect of regener¬ 
ative feedwater heating, if introduced into the steam cycle; the problem, 
in Fig. 14:13 and in the example to follow, has been reduced to its simplest 
proportions. 

The efficiency of this binary-vapor cycle may be found by applying 
Eq. (2:15). Q s includes only heat supplied by the source, or Q s = M M (h d 


hb) — Mm 


hd h a 


{P U M _ PlM^Vg 

~T~ 


where M m is the weight of 


mercury per pound of steam flow, h a and v a are, respectively, the specific 
enthalpy and specific volume of saturated-mercury liquid at the lower 
pressure pi M of the mercury cycle, and p UM is the upper pressure of that 
cycle. The heat rejected to the refrigerator is discharged during the con¬ 
densation of the exhaust steam, and Q R = hj — hf. The efficiency of the 
cycle is therefore 


v = 


Qs Qr 
Q s 


Mm 

hi - h a - j Plu)Va 

-se 

1 

1 

Mm 

^ ( Pum Plu)Va 

J 



(14:14) 


The weight of mercury flow per pound of steam flow may be calculated as 



hj — h g 
he ha 


hi — h f 


(P uw Plw)Vf 



(14:15) 


in which P Uw and P tw refer, respectively, to the upper and lower pressures 
of the steam cycle. 

Example 14:6. In a binary-vapor cycle, mercury is supplied the mercury turbine 
as a saturated vapor at 180 psia. The mercury turbine exhausts to the mercury 
condenser-steam boiler unit at a pressure of 2 psia. The steam leaves this unit dry 
and saturated at 600 psia and is expanded in the steam turbine to a steam-condenser 
pressure of 1 psia. Find (a) the weight of mercury circulated per pound of steam 
flowing around its cycle and (6) the efficiency of the binary-vapor cycle. 

Solution : 

(a) The notation used in this solution will be that of Fig. 14:13. A table giving the 
properties of mercury vapor will be found in the Appendix. Reading from that table, 
















354 


BASIC ENGINEERING THERMODYNAMICS 


hd(h g at 180 psia) = 151.1 Btu; h a ( h / at 2 psia) = 17.65 Btu; hf g at 2 psia = 125.8 
Btu; s e = Sd{s g at 180 psia) = 0.1193; s a (s/ at 2 psia) = 0.02514; s/ g at 2 psia = 
0.1304. 

Then 


s e = 0.1193 = s a + x e (sfg at 2 psia) = 0.02514 + x e (0.1304) or x e = 0.722 
he = ha A X e (h fg at 2 psia) = 17.65 + (0.722)(125.8) = 108.4 Btu 

In the cycle of the water and steam, points i and g correspond to points 2 and 1, 
respectively, of the cycle of Example 14:24.. In that example a calculation was made 
of the difference h 2 — hi = 1131.5 Btu = hi — h 0 . Substituting in Eq. (14:15), 

Mm = iQg 4*— 1 r 7 ~ 6 5 = ^ Hg per pound of H 2 0 


Note that the irreversibilities encountered in the flow of mercury vapor through the 
real mercury turbine would have the effect of increasing h e and thus of considerably 
reducing this ratio. 

(6) The volume v a of the liquid mercury is not shown in the table but, because of the 
high density of liquid mercury, is so small that the work of the mercury-liquid pump 
may be considered negligible. Applying Eq. (14:14), 


12.45(151.1 - 17.65 - 0) - (807 - 69.7) 
12.45(151.1 - 17.65 - 0) 


0.556 


in which hj and h/ correspond, respectively, to h 3 and /i 4 of Example 14:24 and have 
been taken from the solution of that example. The efficiency of the equivalent 


Carnot cycle is 


999.6 - 101.74 
999.6 + 460 


0.615. 


Problems 

1. Sketch, on Ts, hs, and pv coordinates, a Carnot reversible vapor cycle in which 
saturated steam is supplied at 400 psia to the prime mover which exhausts at 1 in. 
Hg abs. (a) What is its efficiency? (5) If the fluid enters the boiler as a liquid, 
what is its quality as it leaves the condenser? (c) What is the quality of the steam 
as it approaches the condenser? ( d) How much heat is transferred, per pound of 
steam, in the boiler? (e) In the condenser? (/) On the basis of the answers to 
parts d and e, check your answer to part a. ( g ) What work is delivered per pound 
of steam by the prime mover? ( h ) How much work must be returned to drive the 
compressor? (i) What are the net work of the cycle and the work ratio? (j) What 
is the steam rate? 

2. Describe the cycle of Prob. 1 as it might be carried out as a series of nonflow 
processes on a pound of H 2 0 confined in a cylinder closed by a piston, (a) What is 
the efficiency? ( b ) What is the work ratio? (c) What is the mep of the cycle? 

3. Introduce into the cycle described in Prob. 1 the assumption that the engine 
efficiency of the prime mover and the adiabatic efficiency of the compressor are both 
70 per cent. Answer the questions asked in Prob. 1. 

4. What is the cycle efficiency in Prob. 1 if (a) the supply pressure is 200 psia? 

(b) If the exhaust pressure is atmospheric ? (c) If both supply and exhaust pressures 

are changed as indicated? 

5. In a Rankine cycle, steam is supplied to the prime mover at the same state 

and the exhaust pressure is the same as for the Carnot cycle of Prob. 1. (a) What 

is the quality of the steam as it approaches the condenser? (6) How much work is 
done per pound of steam by the prime mover? (c) How much work is required to 





POWER—VAPOR SYSTEMS 


355 


drive the feedwater pump? ( d ) What are the net work of the cycle and its work 
ratio? (e) How much heat is transferred, per pound of steam, in the steam generator? 
(/) In the condenser? (g) What is the efficiency of the cycle? ( h ) What is the 
steam rate? (i) Sketch the cycle on Ts, hs, and pv coordinates, (j) Compare your 
answers with those of Prob. 1, and discuss. 

6 . Describe the cycle of Prob. 5 as it might be carried out in a cylinder closed 
by a piston as a series of nonflow processes, (a) What is its efficiency? (6) What 
is the work ratio? (c) What is the mep of the cycle? 

7. Introduce the assumption that the prime mover and the feedwater pump each 
operates with an efficiency of 70 per cent into the cycle of Prob. 5. Answer the 
questions asked in Prob. 5. 

8 . What is the cycle efficiency if the following changes of data are made in Prob. 5? 
(a) The quality of the steam supplied the prime mover is 90 per cent. (b) The 
supply pressure is 200 psia. (c) The exhaust pressure is atmospheric. ( d ) Both 
supply and exhaust pressures are changed as indicated. 

9. The cycle is the same as that of Prob. 5 except that the steam is superheated 
to 700°F before entering the prime mover. Answer the same questions asked in 
Prob. 5. 

10. Assume that the efficiencies of prime mover and feedwater pump are each 
70 per cent in Prob. 9. Answer the same questions. 

11 . Superheated steam is supplied the turbine prime mover at 400 psia. The 
exhaust pressure is 1 in. Hg abs. If the moisture percentage in the exhaust steam 
is not to exceed the maximum suggested in the text, to what temperature must the 
steam be superheated if the efficiency of the prime mover is (a) 70 per cent; (6) 75 per 
cent; (c) 80 per cent; ( d ) 85 per cent; (e) 100 per cent? 

12. Superheated steam at 400 psia and 700°F is supplied to the high-pressure 
turbine of a reheat cycle and expands reversibly until its quality is 0.95. It is then 
extracted and reheated to 700°F before expanding reversibly in the low-pressure 
turbine to 1 in. Hg abs. (a) What is the pressure in the reheater? ( b ) What is the 
quality of the steam as it approaches the condenser? (c) Per pound of steam, how 
much work is performed in the high-pressure turbine? ( d ) In the low-pressure 
turbine? (e) What heat is transferred in the steam generator? (/) In the reheater? 
(g) In the condenser? ( h ) What is the cycle efficiency, and how does it compare 
with that of an equivalent Rankine cycle but without reheat? ( i ) Sketch the cycle 
on hs coordinates. 

13. Steam is to be supplied the high-pressure turbine of a reheat-cycle plant at 
1500 psia, superheated to 800°F. Condenser pressure is 1 in. Hg abs. Based upon 
reheating to 800°F, select a reheating pressure that will give equal and minimum 
percentages of moisture in the steam as it leaves the high- and the low-pressure tur¬ 
bines of the plant. What will that percentage of moisture be if (a) the expansions 
are reversible in both turbines and ( b ) if both have an efficiency of 85 per cent? 
(c) If both efficiencies are 80 per cent? 

14. Work Example 14:4, changing the number of extraction points to two. 

15. Work Example 14:4, changing the number of extraction points to one. 

16. In Example 14:4, assume that the steam is superheated to 800°F before entering 
the turbine. Calculate the same quantities. 

17. In a one-heater regenerative cycle, steam is supplied the turbine as saturated 
steam at 400 psia, and the condenser pressure is 1 in. Hg abs. (a) Select a suitable 
heater pressure. (6) Calculate the efficiency of the cycle, and compare with the 
efficiencies of the equivalent Carnot and Rankine cycles as computed in Probs. 1 
and 5. 


356 


BASIC ENGINEERING THERMODYNAMICS 


18. In a theoretical regenerative vapor cycle, steam is supplied to a 14-stage 
turbine at 500 psia and 640°F. The condenser pressure is 1 in. Hg abs. Assuming 
that the expansion in the turbine is isentropic and that equal work is performed 
in each stage, what are the pressures at the end of each stage? Select extraction 
pressures for (a) a four-heater cycle, (6) a three-heater cycle, to conform to the design 
of the turbine. 

19. In a two-heater regenerative cycle, steam is supplied the turbine at 400 psia, 
700°F. Condenser pressure is 1 in. Hg abs. If the extraction pressures are 100 
and 12 psia, calculate the efficiency and compare with the efficiency of the cycle of 
Prob. 9. 

20. In a combination of the reheat and the regenerative cycles, steam is supplied 
to the high-pressure turbine at 1200 psia and 900°F. After isentropic expansion in 
that unit to a pressure of 90 psia, all of the steam is extracted; a part goes to a regener¬ 
ative heater, and the balance is reheated at constant pressure to a temperature of 
600°F. The reheated portion is then expanded in the low-pressure turbine to con¬ 
denser pressure of 1 in. Hg abs. The condensate is returned to the boiler through the 
regenerative heater by suitable pumps. Draw a flow diagram of this cycle similar 
to Figs. 14:6 and 14:10 and a Ts diagram that combines the features of Figs. 14:7 
and 14:11, lettering the diagrams to correspond with each other. Per pound of steam 
supplied the high-pressure turbine, calculate (a) the weight of steam going to the 
feedwater heater, ( b ) the weight of steam reheated, (c) the heat added in the steam 
generator, ( d ) the heat added in the reheater, (e) the total heat added in the cycle 
from the fuel, (/) the heat rejected in the condenser, and ( g) the total work of the 
cycle. ( h ) What is the efficiency of the cycle? (i) What is its steam rate? Neglect 
pump work throughout. 

21. With respect to the features listed in items 1 to 10 of Art. 14:5, discuss the 
suitability as a vapor for power generation of the following: ( a ) ammonia; ( b ) sulfur 
dioxide; (c) Freon 12; ( d ) carbon dioxide. Would any of these vapors have any 
advantage as one of the vapors in a multivapor power cycle? If so, over what range 
of temperatures would it find use? What advantages would it possess over steam 
in this range? What disadvantages in this range? 

22. In Example 14:6, assume that the efficiency of the mercury prime mover is 
75 per cent and that of the steam prime mover is 80 per cent, (a) What is the ratio 
of the mass rates of flow of mercury and steam if the states of the vapors at entrance 
to the turbines are the same as in the example? (6) What is the efficiency of the 
cycle? (c) What are the qualities of the steam and mercury as they enter their 
respective condensers? 

23. In Example 14:6, assume that the metallurgical limit is 980°F and that the 
pressure in the mercury condenser is 1 psia. A differential of 15°F is maintained 
between the temperature of the condensing mercury and the evaporating steam. 
The steam condenser operates at 1 in. Hg abs. What are the upper pressures of the 
mercury and the steam cycles? Calculate the weight of mercury circulated per 
pound of steam and the efficiency of the binary-vapor cycle. 

Symbols 

c p specific heat at constant pressure 
h specific enthalpy 
J proportionality factor 
M mass rate of flow 
p pressure, psi 

P pressure, psf; pressure in general 
Q rate of heat flow, per cycle or per unit time 


POWER—VAPOR SYSTEMS 


357 


s specific entropy 
T absolute temperature 
v specific volume 
W rate of work delivery 
x quality of the vapor 

Greek Letters 

17 efficiency of the heat engine; thermal efficiency 

Subscripts 

e engine 
i intermediate 
l lower 
L liquid 

M of the mercury (in the binary-vapor cycle) 
p pump; also, at constant pressure 
R refrigerator 
S source 
u upper 
V vapor 

W steam (in binary-vapor cycle) 


CHAPTER 15 


REFRIGERATION 

15:1. Introduction. Refrigeration is defined as the production and 
maintenance of a temperature below that of the surroundings within the 
boundaries of a limited space. To accomplish this effect requires that 
the heat must be continuously removed from the contents of the space 
(called the cold body) and deposited with the surroundings (termed the 
hot body). The refrigerating machine must therefore operate on a cycle 
as it receives heat at a lower and discharges it at a higher temperature; it 
is, in its scheme of operation, a heat pump in the sense in which the use 
of that term was introduced in Art. 4:9 in connection with reversed opera¬ 
tion of the Carnot engine. Although the heat pump may operate as a 
refrigerating machine, we shall make a distinction in the use of the two 
terms that is based on the purposes served as a result of the operation of 
the two devices. The heat pump, we shall say, has the objective of mov¬ 
ing the heat from a lower to a higher temperature level, and the rate of its 
operation is fixed by the rate at which heat is to be discharged at the 
upper temperature; the objective in the operation of the refrigerating 
machine is to maintain a lower temperature in the cold body, and it 
removes only enough heat from that body to accomplish this effect. 

Any reversible engine may operate as a refrigerating machine. In 
operation as an engine, it receives heat at a temperature above that of the 
atmosphere and discharges it at a temperature which may approach that 
of the atmosphere as a lower limit. As a refrigerating machine, on the 
other hand, the surroundings, in practice, usually consist of the atmos¬ 
phere, and the temperature range of operation lies below the temperature 
of the atmosphere; thus the metallurgical limit, which has received so 
much attention in preceding chapters in connection with power cycles, is 
not a factor in the refrigeration cycle. 

In developing the discussion of the refrigeration cycle, it will be our 
policy to consider that the temperatures of both hot body and cold body 
are constant. The atmosphere, which usually constitutes the hot body, 
is vast in extent, and the addition of finite amounts of heat to it will not 
alter its temperature appreciably. The cold body, on the other hand, 
comprises only a limited space, and we should ordinarily expect that the 
continuous removal of heat quantities from it would, ultimately at least, 
lower its temperature. But this cold body is immersed in higher tempera- 

358 


REFRIGERATION 


359 


ture surroundings and simultaneously receives heat from these surround¬ 
ings, as a result of the differential of temperature, as this heat is removed 
by the refrigerating machine; the maintenance of constant temperature 
in the cold space presumes that a delicate balance is maintained between 
the rate at which the refrigerating machine is operated and the rate at 
which heat leaks back into the space from the surroundings. 

The rate at which it is capable of removing heat from the cold body is 
a measure of the capacity of the refrigerating machine, just as the capacity 
of the engine may be expressed in terms of its horsepower. The melting 
of ice is the traditional method of providing refrigeration, and it is not 
surprising to find that the recognized unit of capacity of the refrigerating 
machine is based on the refrigerating effect produced in terms of that 
melting. This unit of capacity is the ton of refrigeration , equivalent in 
refrigerating effect to that produced by the melting of 2000 lb of ice over a 
time period of 24 hr. Since the enthalpy of fusion of ice at atmospheric 
pressure is about 144 Btu, the rate of heat removal from the cold body 
which is defined as a ton of refrigeration amounts to 288,000 Btu in 24 hr, 
12,000 Btu/hr, or 200 Btu/min. 

Another distinction between the refrigeration cycle and the power cycle 
is found in the size of plant to which they are applied. The central- 
station power plant may be designed to meet the requirements of an entire 
area, such as a city or even a state, and large expense for equipment is 
often justified if its efficiency can be thereby increased in even small 
degree. On the other hand, the refrigerating machine is usually a much 
smaller device, and small savings that can be secured only by using com¬ 
plicated and expensive equipment are often ignored in favor of simplicity 
and lower first cost. 

15:2. The Reversed Carnot as a Refrigeration Cycle. The Carnot 
engine cycle has been shown to be the power cycle of highest efficiency. 
As a reversible cycle when the temperatures of source and of sink are con¬ 
stant, it should offer possibilities as a refrigeration cycle which will be 
worth investigating. 

In Fig. 15:1, the Carnot cycle has been plotted as a refrigeration cycle 
on Ts coordinates. It will be observed that the appearance of the cycle is 
unchanged from the form in which, in earlier chapters, we have so often 
had occasion to refer to it; however, the order of cycle processes is now 
different, and the limiting temperatures, denoted formerly by T s and T R , 
are now designated as T H and Tc . Starting from state 1 of the figure, 
with the fluid at the temperature T H of the hot body, the order of cycle 
processes is now as follows: 

1. An isentropic expansion, 1-2, of the working fluid to the temperature 
T c of the cold body. 

2. An isothermal expansion, 2-3, at the temperature T c of the cold 


360 


BASIC ENGINEERING THERMODYNAMICS 


body during which the cold body gives up heat to the working fluid in the 
amount represented by the area 2-3 -b-a of the figure; this may be desig¬ 
nated as Q c in the refrigerating cycle and is analogous to Q R of the power 
cycle except in the direction of its flow. 

3. An isentropic compression, 3-4, of the fluid to the temperature Th of 
the hot body. 

4. An isothermal compression, 4-1, at the temperature T H of the hot 
body. During this process the hot body receives heat from the working 
fluid in the amount represented by the area 1-4-6-a, which may be 
designated as Q H and is analogous to Qs of the power cycle except in 
respect to its direction of flow. Qh is obviously larger than Qc, and since 
the isentropic processes take place without heat flow, their difference 

must be equal to the net work 
which is supplied, by external sys¬ 
tems in the course of the cycle; 
this net work is represented by 
the area 1-2-3-4 enclosed within 
the cycle. 

The cycle of Fig. 15:1 may, as 
we have seen, be traversed as a 
power cycle in a direction opposite 
to that described above; as a 
power cycle its efficiency is 
(T h — Tc)/T h . For a given hot- 
body temperature, this power- 
cycle efficiency could approach 1 
as a limit as T c approaches abso¬ 
lute zero, but in no case could it 
exceed that value, since negative 
temperatures on any absolute scale of temperature have been shown to be 
an impossibility. As a refrigeration cycle, its effectiveness must be 
expressed on a somewhat different basis. The purpose of refrigeration is 
the removal of the heat quantity Q c from the cold body; this is the useful 
effect that should be credited to the refrigerating machine in any attempt 
to rate its performance. The payment made to secure this useful effect 
is in the currency of work, the net work which must be furnished by 
external systems to carry out the cycle. The effectiveness of the Carnot 
refrigeration cycle is therefore expressible as the ratio of the area 2-3 -b-a 
to area 1-4-3-2. It is noted at once that this ratio in Fig. 15:1 is greater 
than 1 and, in fact, could approach infinity as T c approaches T H . The 
ratio is not, therefore, to be classed as an efficiency; instead, the term 
coefficient of performance (cop) is used to express this ratio and thus to rate 
the performance of the Carnot refrigeration cycle and of all refrigerating 



Fig. 15:1. The Carnot refrigeration cycle. 





REFRIGERA TION 


361 


cycles and machines. Even a casual inspection of Fig. 15:1 will indicate 
that, since the Q c and net-work areas are in proportion to their vertical 
heights, the coefficient of performance of the Carnot refrigeration cycle 
is T c /(Th - T c ). 

The principal value of the Carnot power cycle has been found in its 
proved status as a pattern cycle, i.e., a cycle of maximum efficiency at 
given source and refrigerator temperatures. It would seem worth while 
to establish similarly the relative status of the Carnot refrigeration cycle; 
if it can be shown that associated with this cycle is the maximum coeffi¬ 
cient of performance that is conceivable between given cold- and hot- 
body temperatures, the Carnot refrigeration cycle will become an 
important tool for the comparison of the performance of all refrigeration 
cycles and refrigerating machines with the ideal. To accomplish this 
purpose, let us assume, in a manner that is reminiscent of the method of 
proof of the Carnot principle in Chap. 5, that a refrigerating machine 
(working on a cycle and thus capable of continuous operation) is available 
which has a higher coefficient of performance than that of the Carnot 
when placed between cold and hot bodies at temperatures T c and T H , 
respectively. This would mean that, at the expenditure of the same 
amount of work that must be furnished by external systems to operate 
the Carnot refrigerating machine, a greater amount of heat Qc could be 
removed from the cold body by this hypothetical machine and the heat 
Q h delivered to the hot body would also be larger than in the operation 
of the Carnot machine. This heat Q H could then furnish the heat 
supply to a Carnot engine, which would exhaust heat to the cold body. 
The work output of this Carnot engine would be greater than that 
required to drive the hypothetical machine, and since the hot body has 
given and received the same amount of heat and the only net heat 
exchange has therefore been with the cold bodjq a violation of the Second 
Law is evident. The assumption that a refrigerating machine may be 
devised that has a higher coefficient of performance than the Carnot when 
operating between the same limiting temperatures must therefore be 
abandoned. Thus the- ideal, or maximum conceivable, coefficient of 
performance is that of the Carnot refrigeration cycle, or 


COPideal 



(15:1) 


The coefficients of performance of refrigeration cycles and refrigerating 
machines will vary over a wide range according to the ratio T C /T H . 
The mere fact that a cycle or a machine shows a high cop is not an 
indication that no considerable improvement is possible; nor does a low 
cop necessarily mean that effort spent in an attempt to increase the per¬ 
formance will be rewarding, for this apparently low value may still be a 



362 


BASIC ENGINEERING THERMODYNAMICS 


large fraction of the ideal cop if T c is low in comparison with T H . To 
express the ratio of the performance of the refrigerating cycle or machine 
to the ideal standard set by the Carnot refrigeration cycle, the term 
refrigerating efficiency is used. Efficiency in this case does not represent 
a ratio of output to input, as for the efficiency of a heat engine but, like 
the engine efficiency discussed in Chap. 13, is instead a ratio of actual to 
ideal performance. It may be expressed as the ratio of the actual cop to 
the ideal cop, or 


cop 

rjr 

COPideal 


T h - T c 
cop- T - 

i c 


(15:2) 


The working substance in a refrigerating machine is called the refriger¬ 
ant. When the refrigerant is a gas or a superheated vapor, the Carnot 
cycle suffers from the same or even greater handicaps as a refrigeration 

than as a power cycle, owing to 
the impracticability of heat transfer 
at constant temperature and the 
low work ratio of the cycle. For 
there was an advantage in using a 
gas as the working substance in the 
power cycle if internal combustion 
could thereby be employed and the 
temperature T s increased above the 
metallurgical limit, while, in the 
refrigeration cycle, the range of 
temperatures falls, in practice, en¬ 
tirely below this limit. The use of 
a saturated vapor and a steady- 
flow cycle makes possible the trans¬ 
fer of heat simultaneously at con¬ 
stant temperature and at constant 
pressure, a practical method of 



S 


Fig. 15:2. The Carnot refrigeration cycle 
for a saturated vapor. 


steady-flow heat exchange. Moreover, the work ratio of the cycle is 
increased, and it is therefore considerably less vulnerable to the effects 
of friction and other irreversibilities in the real plant. Although air was 
formerly used as a refrigerant in some special situations, the vapor has 
now completely replaced the gas for the purpose. 


In Fig. 15:2, the Carnot cycle of Fig. 15:1 has been placed between the 
saturated-liquid and saturated-vapor lines of a suitable refrigerant. The 
general description of the cycle is, however, the same as was outlined near 
the beginning of this article. I he apparatus necessary for a demonstra¬ 
tion of this cycle is shown in the form of a flow diagram in Fig. 15:3. The 
vapor engine accepts the saturated liquid at the upper pressure p u and 







REFRIGERA TION 


363 


upper temperature T H of the cycle and expands it isentropically to the 
lower pressure p t and lower temperature T c . By applying Eq. (3:5), the 
work delivered by this unit may be shown to equal hi — h 2 , per pound of 
refrigerant; this (positive) work is represented in Fig. 15:2 by the triangu¬ 
lar area c-1-2. The refrigerant enters the evaporator as a wet vapor with 
a large percentage of moisture, as shown by the position of point 2 of Fig. 
15:2. This evaporator is located in the cold room, so called because the 
purpose of operation of the machine is to maintain this space at low tem¬ 
perature. As the cold room receives heat from its surroundings, which 
are at higher temperature, it passes this heat on to the refrigerant as the 



Fig. 15:3. Flow diagram—Carnot vapor-refrigeration cycle. 

latter flows through the evaporator, with the result that the liquid gradu¬ 
ally evaporates until the mixture leaves the evaporator at state 3 of Fig. 
15:2. This represents the steady-flow transfer of heat in the amount 
hz — h 2 , per pound of refrigerant, according to Eq. (3:5). 

Entering the vapor compressor at state 3, the vapor is compressed isen¬ 
tropically to p u and T h . In Fig. 15:2, state 4 at exit from the compressor 
has been shown as a dry and saturated state, but both points 3 and 4 could 
move to the left (though not to the right) without affecting either the 
coefficient of performance of the cycle or the possibility of carrying out the 
steady-flow heat exchange in the condenser as a constant-pressure process. 
This is called wet compression since it is carried out entirely within the 
saturated-vapor region. The work of compression, from Eq. (3:5), is 
h\ — hz, per pound of refrigerant, and this work is represented by the 
trapezoidal area c-1-4-3. Taking into consideration the positive work of 
the engine, the net work of the cycle is represented by the rectangular area 




















































364 


BASIC ENGINEERING • THERMODYNAMICS 


1-2-3-4. The work ratio of the cycle is the ratio of this net-work area to 
the area c- 1-4-3, and a comparison of these areas shows the work ratio to 
be high and therefore favorable. The last process of the cycle takes place 
in the condenser as the vapor is condensed to a saturated liquid at con¬ 
stant temperature T H and constant pressure p u . The heat rejected to the 
hot body in this steady-flow heat exchanger is h\ — hi, per pound 
condensed. 


Example 15:2. A Carnot refrigeration cycle, using ammonia as the refrigerant, 
operates between a cold-room temperature of 0°F and atmospheric temperature of 
90°F. (a) What is the cop of this cycle? ( 6 ) Calculate, based on the properties 

of ammonia, the heat removed from the cold body, the compressor work, engine work, 
and net work of the cycle per pound of ammonia flow, (c) Calculate, per ton of 
refrigeration, the mass flow rate, the volume of ammonia per minute handled by the 
compressor, and the power required, (d) What is the pressure ratio of compression? 

Solution. The properties of ammonia may be taken from the abbreviated table 
that will be found in the Appendix. 

(a) cop Th _ ^ 90 _ o - 5.11 

(1 b ) hi = hf at 90°F (180.6 psia) = 143.5; a, = 0.2958 
hi = h g at 90°F = 632.0; s 4 = 1.1846 

«2 = Si = 0.2958 = s / 2 -f- x 2 Sfy 2 = 0.0975 x 2 (1.2377) or x 2 = 0.16 

h 2 = hf 2 + x 2 h f g 2 = 42.9 + (0.16) (568.9) = 133.9 Btu 

s 3 = s 4 = 1.1846 = 8 ft + x 3 s /„ 3 = 0.0975 + x«(1.2377) or x 3 = 0.878 

hz = hf 3 + xzh f0z = 42.9 + (0.878) (568.9) = 542.5 Btu 

Qc = hz - h 2 = 542.5 - 133.9 = 408.6 Btu/lb 
W 

~ = h* - h 3 = 632.0 - 542.5 = 89.5 Btu /lb 
W 

-j = hi - h 2 = 143.5 - 133.9 = 9.6 Btu/lb 
Net work supplied = 89.5 — 9.6 = 79.9 Btu/lb 


Checking, 

JQc 408.6 
C0P IF net 79.9 


5.11 


This agrees with the answer to part a. 


(c) 


200 _ 200 
Qc 408.6 


V = Mv 3 = MxzV g 


0.49 lb NH 3 per minute per ton 
= (0.49) (0.878) (9.116) = 3.92 ft 3 /(min) (ton) 


= (200) (778) = (200) (778) 

P cop(33,000) (5.11) (33,000) 

(d) Pressure ratio = — 4 = = 5.93 

Pz 30.42 


0.92 hp/ton 


15:3. The Vapor-compression cycle is a variant of the Carnot refrigera¬ 
tion cycle in which certain changes are introduced with the purpose of 
simplifying the apparatus needed for the refrigeration plant and making 
it easier to operate and maintain. 












REFRIGERATION 


365 


One change from the Carnot cycle which is always made in the real 
vapor-compression plant is the substitution of an expansion valve for the 
vapor engine of Fig. 15:3. Even if isentropic expansion in the engine were 
possible, the work delivered by this unit has been seen above to be rela¬ 
tively small. The irreversibilities which will always accompany its 
operation as a real device will still further reduce the amount of work 
delivery. The expansion valve is a simple de\ r ice that amounts to a valve 
with an opening for the flow of refrigerant which can be enlarged or 
reduced in size according to the pressure on its downstream side. Thus 
the flow of refrigerant may be controlled automatically, increasing as the 
pressure in the evaporator falls and 
decreasing as this pressure rises. t 

The use of an expansion valve 
substitutes an irreversible adiabatic 
expansion for the reversible adia¬ 
batic process 1-2 of Fig. 15:2. 

This irreversible adiabatic process 
is of a throttling character and is 
similar to the process carried out in 
the throttling calorimeter and dis¬ 
cussed in Art. 7:10. Thus = hi, 
and state 2 moves horizontally to 
the right as indicated in Fig. 15:4. 

We should expect to find that the 
introduction of this irreversibility 
into the cycle has had the effect of ^ 
lowering its coefficient of perform- ^ 
ance. This reduction in the coef¬ 
ficient of performance results both from the reduction in Qc which accom¬ 
panies the movement of state 2 to the right and irom an increase in the 
net work of the cycle caused by the elimination of the work return for¬ 
merly made by the vapor engine. The net work of the cycle is now the 
work of the vapor compressor, or /i 4 — hz per pound. 

Another change from the Carnot refrigeration cycle that is usually 
found in the vapor-compression cycle consists in allowing the evaporation 
of the vapor in the cold room to continue until the vapoi becomes diy, in 
Fig. 15:4, this alteration is shown in the shifting of point 3 to the right 
from e, which is its position in the Carnot cycle of Fig. 15:2. This causes 
state 4 at exit from the compressor to lie in the superheated-vapoi legion 
and is called dry compression to distinguish it from the wet-compiession 
process of the Carnot cycle. Although it introduces a second irreversi¬ 
bility into the cycle and thus still further reduces the refrigerating effi¬ 
ciency, dry compression is usually preferred because of simplifications of 



. 15:4. The vapor-compression cycle— 
compression. 









3G6 


BASIC ENGINEERING THERMODYNAMICS 


operation and control that are made possible in the real refrigerating 
machine. The vapor gives no readily observable signal as it approaches 
and passes point e of Fig. 15:4 in the course of its evaporation; neither its 
pressure nor its temperature undergoes any change along the entire path 
2-3. Thus no indication is given either a human operator or an auto¬ 
matic-control apparatus that would aid in the operation of the compressor 
in such a manner or at such a rate that wet compression will be effected. 
On the other hand, if evaporation is allowed to continue to point 3 of Fig. 
15:4, the passing of this point will be signaled by a rise in the temperature 
of the refrigerant as it begins to acquire superheat. 

Dry compression has what appears to be a secondary practical advan¬ 
tage in that the Qc area is increased as point 3 is shifted to the right (see 
area A Q c of Fig. 15:4), and thus the refrigerating effect is larger per pound 
of refrigerant circulated. This advantage is, however, less than it might 
appear, for the size of the compressor is determined by the volume of 
refrigerant it must handle and the specific volume at state 3 is greater 
than at state e of Fig. 15:4; also, heat exchange in both evaporator and 
condenser is greater per pound of refrigerant passing through those units, 
and their design must take this into consideration. Thermodynamically, 
A Q c is expensively obtained, for it requires the extra work area d- 4-3-c of 
the figure. The coefficient of performance associated with the removal 
of A Q c from the cold body is therefore represented graphically in Fig. 

area e-3-b-f 


15:4 as the ratio 


This ratio is less than the ideal cop of 


area d-4-3-e 

Tc/(T h — T c ), as is clear from a visual inspection of the proportions of 
the diagram, and it is evident that dry compression results in lowering the 
cop of the entire cycle. The irreversibility which is the thermodynamic 
explanation of this loss in performance consists in the “downhill” flow of 
heat during the desuperheating process 4 -d ; the coolant must be supplied 
at the temperature T H or below if the later condensation d -1 is to be 
possible. 

In the real refrigeration plant, the temperature of the coolant must be 
somewhat less than the T H of the cycle and the temperature of the cold 
room somewhat above that of the refrigerant in the evaporator if the heat 
exchange in the condenser and evaporator, respectively, is to proceed at 
the desired rate. These irreversibilities, combined with others encoun¬ 
tered in the course of the circuit (principally in the compressor), cause the 
cop of the real plant to be less than that of the cycle on which its operation 
is based. The compressor of the real plant may be of either the positive- 
displacement or the centrifugal type, according to the volume of refriger¬ 
ant it is required to handle and the differential of pressure, p u — pi. 

The flow diagram of the vapor-compression cycle is presented in Fig. 
15:5. The coefficient of performance of the cycle is 



REFRIGERA TION 


367 


cop = 


J Qc 

W~ c 


hz /12 

hi hz 


hz — hi 
hi — hz 


hz — hf u 
hi hz 


(15:3) 


in which hi (= /i 2 ) is the enthalpy of the refrigerant as a saturated liquid 
at condenser pressure p xl and hz and hi are, respectively, the specific 
enthalpies of the refrigerant at the beginning and the end of the isentropic 
compression. 1 

Example 15:3. For a vapor-compression refrigerating cycle using ammonia and 
operating between a cold-room temperature of 0°F and an atmospheric temperature 
of 90 F, calculate the coefficient of performance, the refrigerating efficiency, the heat 



Fig. 15:5. Flow diagram—vapor-compression cycle. 

removed from the cold body and the net work of the cycle per pound of refrigerant, 
and, per ton of refrigeration, the mass flow rate, the volume per minute handled by 
the compressor, and the horsepower required, for (a) wet compression, and (6) dry 
compression. 

Solution: 


(a) hi = 143.5; hz = 542.5; h 4 = 632.0; Xz = 0.878 (see Example 15:2); h 2 = hi 
= 143.5. From Eq. (15:3), 


cop 


V r 


542.5 
632.0 
4.46 
5.11 : 


- 143.5 

- 542.5 

= 0.873 


399.0 

89.5 


4.46 


The substitution of an expansion valve for the vapor engine of the Carnot cycle 
accounts for this reduction of 12.7 per cent in performance. 

1 If wet compression is employed, point 3 shifts to position e of Fig. 15:4 and point 

4 to d. 














































368 


BASIC ENGINEERING THERMODYNAMICS 


Qc = hz - h 2 = 542.5 - 143.5 = 399.0 Btu/lb 

Net work of cycle = = hi — h 3 = 632.0 — 542.5 = 89.5 Btu/lb 


M = = ggg-Q = 0.501 lb/(min) (ton) 

V = Mx 3 v 0s = (0.501) (0.878) (9.116) = 4.01 ft 3 /(min)(ton) 


(200) (778) (200) (778) 

hp cop(33,000) (4.46) (33,000) 


1.06 hp/ton 


( 6 ) Use is made of the table of properties of superheated ammonia to find h \: s 4 = S 3 
= s g at 0°F = 1.3352; p 4 = saturated pressure at 90°F = 180.6 psia. Interpolation 
in the table based on this pressure and entropy gives = 724.5. 


hz = h g at 0°F = 611.8; h 2 = hi 


cop 


Vr 


611.8 - 143.5 


724.5 

4.16 

5.11 


- 611.8 
= 0.815 


468.3 

112.7 


= 143.5 
4.16 


Qc =h 3 - h 2 = 611.8 - 143.5 = 468.3 Btu/lb 
Net work of cycle — hi — hz = 724.5 — 611.8 = 112.7 Btu/lb 


M 

V 

hp 


200 


200 


= 0.427 lb/(min)(ton) 


Qc 468.3 
Mvz = (0.427)(9.116) = 3.90 ft 3 /(min)(ton) 
(200) (778) 


(4.16) (33,000) 


= 1.13 hp/ton 


15:4. Refrigerants. Any refrigerant, if put through the succession 
of processes which we know as the Carnot refrigeration cycle, would yield 
the maximum coefficient of performance that is associated with that cycle. 
The choice of a refrigerant must therefore be made on the basis of how 
closely the processes of the real cycle, when applied to the refrigerant, may 
be made to approach those of the Carnot. Thus we have already had 
occasion to observe that the vapor has an overwhelming advantage over 
the gas if the cycle operations can be confined to handling the vapor in its 
saturated state. The choice of a suitable refrigerant consists of the selec¬ 
tion of a vapor that will best meet the requirements. 

In the real refrigeration cycle, the range of temperatures is between the 
upper limit fixed by the temperature of the atmosphere and a lower limit 
that is determined by the temperature which it is desired to maintain in 
the space to be cooled. This lower limit of temperature may be all the 
way from temperatures only slightly below that of the atmosphere down 
to temperatures as close to absolute zero as can be attained, and the selec¬ 
tion of the vapor that is to be the refrigerant will depend upon the range 
of temperature through which its use is planned. When the over-all 
range of temperature is very large, it may be desirable, or even necessary, 
to apply the multivapor principle to refrigeration, in much the same man¬ 
ner as it has been applied in the mercury-steam binary-vapor cycle to the 













REFRIGERA TION 


369 


power cycle. This is not done in ordinary commercial refrigeration, 
however. 

Steam is preponderantly the vapor selected as the working fluid of the 
vapor power plant; in refrigeration a much wider selection of vapors is 
available, and resort may even be made to chemical synthesis, as in the 
case of the Freon series of refrigerants, to widen the field of selection. 
The desirable characteristics of the ideal refrigerant include: 

1. It should be cheap, stable, nonexplosive, noncorrosive, and non- 
poisonous and should not dissolve any of the materials which are used to 
confine it in the course of the cycle. 

2. Its critical temperature should be well above the upper temperature 
of the refrigerating cycle so that condensation may take place, at least 
largely, at constant temperature. 

3. Its saturation pressure at the upper temperature of the cycle should 
be moderate to prevent high stresses in the metal of the condenser and 
thus to reduce design difficulties and equipment costs. 

4. Its enthalpy of vaporization should be large as compared with the 
specific heat of the liquid. The basis of this specification will be clarified 
when reference is made to Fig. 15:2. A low specific heat of the liquid 
would be reflected in greater steepness of the saturated-liquid line between 
c and 1 of that figure, and the work of the vapor engine of the Carnot cycle 
would be correspondingly less. When a shift is made to the expansion 
valve of the vapor-compression cycle of Fig. 15:4, the irreversibility of the 
latter cycle would be lessened and the movement of point 2 to the right of 
its position in the Carnot cycle would be shortened, resulting in a smaller 
reduction in Q c as well as a smaller increase in the net work of the cycle. 
As a result, the refrigerating efficiency of the vapor-compression cycle is 
increased. 

5. Its triple-point temperature should be well below the lowest tem¬ 
perature to be reached in the course of the cycle in order that there will 
be no chance of the formation of a solid. 

6. Its saturation pressure at the lower temperature of the cycle should 
be slightly above atmospheric pressure. This provision both decreases 
the volumes that must be handled by the compressor and ensures that air 
will be prevented from leaking into the system. 

7. The refrigerant, as a liquid, should wet the metal walls with which it 
comes in contact in the condenser and evaporator. This permits a reduc¬ 
tion of temperature differential and of heat-transfer surface in these units 
by encouraging high rates of heat transfer. 

8. Its entropy, as a saturated vapor, should be slightly less at the lower 
temperature of the cycle than at the higher. Allowing for the irreversi¬ 
bilities of the compressor, this would permit wet compression, even when 


370 


BASIC ENGINEERING THERMODYNAMICS 


compression started from a dry saturated-vapor state such as point 3 of 
Fig. 15:4. 

No refrigerant possesses all of the properties listed above. The particu¬ 
lar refrigerant to be used is selected with a view to its suitability over the 
temperature range and under the conditions of operation that apply in the 
proposed plant. Water, the almost universally used fluid in the vapor 
power plant, is sometimes used as a refrigerant, but its use, because of 
grave shortcomings with respect to items 5 and 6, is confined to situations, 
as in air-conditioning installations, where evaporator temperatures below 
about 40°F are not required. Even at this temperature, tremendous vol¬ 
umes must be handled, and an ejector type of vapor pump is used as the 
compressor. Other, more commonly used refrigerants include ammonia, 
carbon dioxide, Freon 12, sulfur dioxide, methyl chloride, and propane. 
When the refrigerating efficiency of the cycle of Fig. 15:4 is calculated for 
these vapors over a typical temperature range, they all, with the exception 
of carbon dioxide, show a value of about 80 to 85 per cent. The refriger¬ 
ating efficiency, when carbon dioxide is the vapor, is much lower, being 
less than 50 per cent, but carbon dioxide is none the less used to some 
extent on ships and in other situations where its nontoxic properties are of 
maximum importance; the Freon series of refrigerants are gradually 
replacing carbon dioxide, however, since they enjoy its advantages and, 
at the same time, make possible higher coefficients of performance. 
Ammonia is undoubtedly the most widely used refrigerant, the preference 
given it being based on items 2, 3, 4, 5, 6, and 7 above. 

15:5. Absorption refrigeration is based on a cycle of the refrigerant 
very similar to that of Fig. 15:4 but substitutes a method of passing from 
state 3 to state 4 that does not require a vapor compressor. The change 
is made possible by the ability of some vapors to dissolve in certain sol¬ 
vents; in the commercial absorption-refrigeration plant the vapor is 
usually ammonia, and the solvent is water. 

When anhydrous ammonia (NH 3 ) is dissolved in water, a chemical 
change to NH 4 OH takes place and heat is released as a result. The solu¬ 
tion of ammonia in water is called aqua ammonia , and the concentration 
of ammonia in such solutions is expressed as a ratio of the weight of anhy¬ 
drous ammonia which has been dissolved to the total weight of the solu¬ 
tion; it is represented by the symbol x. When the concentration reaches 
a certain maximum, the water will refuse to accept any more ammonia in 
solution. This maximum concentration depends upon the pressure and 
the temperature of the solution, increasing with increased pressure and 
lowering as the result of higher solution temperature. When ammonia is 
dissolved in water at constant pressure, the heat release that accompanies 
the solution process will cause the solution temperature to rise, the point 
at which no more ammonia can be dissolved will soon be reached, and the 


REFRIGERATION 


371 


process will come to a halt. In order to make it possible for the solution 
to take place continuously, two steps must be taken: provision must be 
made for the continuous removal of heat from the solution so that its tem¬ 
perature will not rise above a permissible level, and some of the strong 
solution that is accumulating must be removed and replaced by fresh sup¬ 
plies of Avater or, at least, by aqua ammonia of lower concentration. 

When it is desired to drive the ammonia out of solution at constant 
pressure, the temperature is raised to a level at which the aqua ammonia 
can no longer retain so large a proportion of ammonia in solution. To 
maintain this temperature as ammonia is driven off in a continuous 
process requires that heat be supplied continuously, since this chemical 


Reefi fie r 



Fig. 15:6. Flow diagram—ammonia-absorption refrigeration cycle. 


reduction of ammonia to the anhydrous form is endothermic, i.e., requires 
the addition of heat from external systems if the temperature of the sys¬ 
tem is not to fall. As a result of the addition of this heat, certain amounts 
of water will also change to vapor form and escape into the space above 
the aqua ammonia, but the proportion of ammonia in the vapors above is 
much higher than in the liquid below. The small amount of water vapor 
in the vapor mixture can be largely reduced to liquid phase and returned 
to become again a part of the aqua ammonia if the vapors, after they have 
been transported to a location Avhere they are no longer in communication 
with the aqua ammonia, are cooled to lower temperature. 

The flow diagram of the ammonia-absorption refrigerating cycle is 
shown in Fig. 15:6. It will be observed that the apparatus concerned in 
the demonstration of processes 4-1, 1-2, and 2-3 remains the same as for 
the vapor-compression cycle of Fig. 15:5. In order to carry out process 




















































































372 


BASIC ENGINEERING THERMODYNAMICS 


3-4 of the vapor-compression cycle, equipment consisting of an absorber, 
a liquid pump, a generator or still, and a rectifier, or drier, replaces the 
vapor compressor of that cycle. 

The dry anhydrous ammonia vapor leaving the evaporator at state 3 
flows to the absorber, where it is allowed to bubble up through, and be 
absorbed in, the water contained in that unit. In order to make partial 
provision for continuous operation, heat is continuously removed from 
the aqua ammonia formed by this solution process at a rate sufficient to 
keep the temperature constant. This heat goes into cooling-water coils 
and thus is discharged ultimately to the atmosphere at the temperature 
level T h of the vapor-compression cycle. The pressure in the absorber is 
approximately the same as that in the evaporator, since an open line con¬ 
nects the two. The temperature of the aqua ammonia in the absorber is 
only enough above T H so that heat may be discharged at the required 
rate; this rate is designated as Q a in the figure. 

By cooling the aqua ammonia in the absorber, only partial provision 
has been made for continuous operation; in addition, aqua ammonia of 
high concentration must be removed and replaced by a weaker solution. 
Thus a liquid pump is provided which draws the solution from the absorber 
and discharges it into the generator. The pressure in the generator is the 
same as that of the ammonia in the condenser and thus is the higher pres¬ 
sure p u of the vapor-compression cycle; the pump is therefore necessary to 
permit the passage of the liquid from absorber to generator against the 
differential between the pressures in the two units. It is in the generator 
that the ammonia is driven out of solution so that it alone can continue 
through the balance of the cycle. In order to do this, the temperature of 
the aqua ammonia must be raised, and heat must be continuously sup¬ 
plied. Since the temperature of the solution is already at or above that 
of the atmosphere, this heat Q g must be supplied at some higher tempera¬ 
ture, which we shall designate as T s ', in the real ammonia-absorption 
plant, the temperature of the steam in the steam coils within the generator 
that furnish this heat is about 300°F. Because of the high temperature 
in the generator, the maximum concentration that can be maintained in 
the aqua ammonia which occupies the lower part of that unit is, in spite of 
the higher pressure in the generator, less than is maintained in the 
absorber, and ammonia vapor (with some water vapor) is driven out of 
solution and into the space above. The weaker solution that remains is 
drawn off and returned to the absorber, thus completing the requirements 
for continuous operation of both absorber and generator. No pump is 
necessary on this return line because the flow is in the direction of lower 
pressure, but a float-controlled valve ensures that the liquid levels in both 
generator and absorber will remain constant. The temperature of the 
liquid in this return line, to be later cooled in the absorber, is much higher 


REFRIGERA TION 


373 


than that of the liquid in the line from absorber to generator, which is to 
be later heated in the generator, and the real absorption plant always uses 
a heat exchanger which brings the two lines into communication and 
allows the temperature of the liquid in the return line to be lowered and 
that in the other line to be raised. This is a feature which is not essential 
to the thermodynamic demonstration of the cycle and has been omitted 
from the figure to avoid possible confusion. 

The ammonia vapor in the upper part of the generator is now ready to 
proceed through the balance of the cycle except that a small percentage of 
water vapor is mixed with it. To eliminate this water vapor, as far as 
possible, the vapor mixture is drawn out of the generator and passed 
through a rectifier, where it is brought into contact with cooling-water 
coils. The temperature of the mixture falls, as a result, and a large pro¬ 
portion of the water-vapor content is condensed and returned to the 
generator as a liquid. The rest of the vapor, now containing only a very 
small percentage of water vapor, continues its flow through the ammonia 
condenser, expansion valve, and evaporator to complete the cycle. 

The advantage of the absorption over the vapor-compression system of 
refrigeration is found in the smaller amount of energy in the form of work 
that must be supplied by external systems to accomplish a given refrig- 
erative effect. It has been shown previously that the work required for 
continuous compression may be measured as JV dP, in which V is the 
volume of fluid compressed, or pumped, and dP is the differential of pres¬ 
sure. The differential of pressure is the same for both cycles, but the vol¬ 
ume of vapor that must be handled by the vapor compressor in order to 
pass a unit weight of refrigerant around the cycle is very much greater 
than the volume of aqua ammonia handled by the pump of the absorption 
cycle in producing the same refrigerative effect. This is true in spite of 
the fact that the pump must handle a considerably greater weight of fluid 
than the compressor and results from the very small specific volumes of 
liquids in comparison with the specific volumes of their vapors. On the 
other hand, large amounts of energy in the form of heat, which must be 
supplied at a temperature well above that of the atmosphere, are required 
in the course of the absorption cycle, and this extra heat must ultimately 
be discharged to the atmosphere at the expense of the use of larger quanti¬ 
ties of cooling water. The absorption plant is less flexible under variable¬ 
load conditions of operation and is more complex and expensive. The 
choice between the two systems should be based on the relative costs of 
energy in the form of heat and as work and upon the conditions of opera¬ 
tion that are anticipated. 

Equation (15:3) may not properly be used for the calculation of the 
coefficient of performance of the absorption refrigeration plant. The 
basic definition of the coefficient of performance of a refrigerating plant 


374 


BASIC ENGINEERING THERMODYNAMICS 


or cycle is the ratio of the heat removed from the space to be cooled 
divided by the work which must be supplied by external systems to 
accomplish that refrigerative effect. If “ work ” in this definition is taken 
literally, the work of the liquid pump of the absorption cycle is so small 
that coefficients that are unreasonably high are obtained, values that 
would exceed the standard set by the Carnot refrigeration cycle by a wide 
margin. This basis of calculation would ignore the large amount of 
energy in the form of heat that must be supplied at a temperature above 
that of the atmosphere to carry out the absorption cycle. In the vapor- 
compression cycle, we are basically concerned with only two levels of 
temperature; these are T c , the temperature of the cold room, and T H , the 
temperature of the atmosphere. In the absorption cycle of Fig. 15:6, a 
third temperature T s is added at which the heat Q g is supplied at the 
generator. If this amount of heat had been furnished a Carnot engine 
which exhausted to the temperature of the atmosphere, an amount of 
work equal to Q g (T s — T H )/T S could have been made available. Con¬ 
structively, if the negligible amount of work actually required to drive the 
liquid pump is ignored, this is the amount of work that should be charged 
to the operation of the absorption plant in calculating its coefficient of 
performance, or 


COPabsorption cycle 


QcT 


s 


Q g (T s - T h ) 


(15:4) 


When the coefficient of performance of the absorption plant is calculated 
on this basis, it cannot exceed the ideal set by the C.arnot cycle and, in 
practice, will fall considerably below the cop of an equivalent vapor- 
compression plant. 

Although the liquid pump of the ammonia-absorption cycle described 
above need be supplied work in only negligible quantities, it must be a 
positive-displacement device with moving parts because of the large 
differential of pressure between absorber and generator. The Electrolux 
refrigerator applies the principle of partial pressures to do away with this 
requirement. It was brought out in Chap. 10 that the temperature at 
which a liquid will evaporate into the space above will depend not upon 
the total pressure of all the gases that occupy that space but only on the 
partial pressure of its own vapor in the space. 

If the temperature in the evaporator coils of the ammonia-absorption 
system is 0°F, the pressure of the saturated ammonia vapor is, according 
to tables of properties of ammonia, about 30 psia. In the condenser, an 
ammonia temperature of 90°F corresponds to a vapor pressure of about 
180 psia; it is against a differential of pressure of this order that the liquid 
pump of the system would operate if ammonia alone were contained in the 
system. If, however, the pressure in the evaporator and absorber can be 



REFRIGERA TION 


375 


built up to equal that in the generator and condenser by adding another 
gas in the evaporator and absorber in an amount sufficient so that its 
partial pressure will make up the difference between the ammonia-vapor 
pressures in the two parts of the system, a balance of total pressure around 
the circuit will be secured. The pump is concerned only with differences 
in total pressure and, in the case where that total pressure is balanced, need 
create only the small pressure differential necessary to overcome the fric¬ 
tional resistance to flow around the circuit at the desired rate. This 
frictional resistance is small if velocities are kept low and, in the Electro¬ 
lux refrigerator, is overcome by allowing the aqua ammonia to enter the 
base of a vertical tube while heat is being supplied to it as it enters the 
generator. This heat causes the entering solution to release ammonia 
vapor, and the action is similar to the percolator of a coffeepot, the metal 
piston required in the mechanical pump being replaced by globules of 
vapor which force slugs of liquid ahead of them as they rise in the tube, 
thus overcoming the small total-pressure difference. 

Hydrogen gas is used to build up the pressure in the evaporator and 
absorber of the Electrolux machine. The choice of hydrogen is based on 
its insolubility in water. It does not go into solution in the absorber and 
thus is not carried into the generator; a liquid seal is also used at the 
expansion valve to prevent it from backing up into the condenser, where 
its pressure would destroy the delicate balance of pressures that is required 
in the system. 

The size of system to which the Electrolux principle may be applied is 
somewhat limited since a large system would offer resistances to flow too 
large to be handled by the percolator type of pump. In the past, its 
application has been limited to small household refrigerators, using a 
small gas flame to supply heat to the generator. More recently, its field 
of use has been widened to include the refrigerative air conditioning of 
buildings of limited size. In the latter application, it is of course neces¬ 
sary that the heat removed to the atmosphere from absorber and con¬ 
denser be voided outside the space which is to be cooled. 

15:6. Adsorption refrigeration is based on the property of some solids 
of attracting the vapor of a surrounding atmosphere and causing the 
formation of a dense layer of the vapor on the surface of the solid. The 
process of adsorption is accompanied by the release of heat, and the tem¬ 
perature of the adsorbing material will rise above the temperature of the 
atmosphere from which the vapor has been adsorbed. The maximum 
amount of vapor which may be adsorbed increases with decreased tem¬ 
perature and with increased pressure. 

In a refrigerator that is based on the adsorptive principle, the solid is 
silica gel and the vapor sulfur dioxide. Two bodies of silica gel are pro¬ 
vided; one is confined in a space at low pressure and adsorbs sulfur dioxide 


376 


BASIC ENGINEERING THERMODYNAMICS 


at low temperature while the other, already charged with sulfur dioxide, 
is located in another compartment and gives off sulfur dioxide vapor as it 
receives heat. The vapor driven off in the high-pressure compartment 
passes through a condenser and is condensed; it then expands through an 
expansion valve into the low-pressure compartment and creates a low 
temperature in that space. The pressure in the low-pressure low-tem¬ 
perature compartment builds up only slowly because of the adsorption 
that is taking place in that space; in the meantime, refrigeration is pro¬ 
vided by the evaporation of the liquid sulfur dioxide after it passes the 
expansion valve. Ultimately the process as described must come to a 
halt as the silica gel in the warm compartment gives up its sulfur dioxide 
content and that on the cold side of the expansion valve adsorbs it. dhe 
position of the two compartments is then reversed, that containing the 
silica gel which has been voided of sulfur dioxide being placed in the space 
to be refrigerated and the other heated so that the vapor held by its gel 
will be driven off. The pressure in the latter compartment builds up as 
the sulfur dioxide is driven off while, on the other side, the pressure drops 
as the gel begins to adsorb sulfur dioxide due to the lower temperature to 
which it is exposed. By means of a suitable valve arrangement, the 
condenser is placed on the opposite side of the expansion valve, and the 
direction of fluid flow is reversed. The commercial application of adsorp¬ 
tion refrigeration is limited because of the intermittent character of its 
operation. 

A schematic diagram of the flow which is described above is shown in 
Fig. 15:7. In the valve position shown, silica gel chamber C 1 is being 
heated and the sulfur dioxide leaves C\ at high pressure and passes through 
the condenser and the expansion valve into C 2 . In this second chamber, 
which is placed in the cold room, the silica gel is adsorbing the vapor. 
When the direction of flow is to be reversed, the valve is given a quarter 
turn clockwise and C\ is placed in the cold room while C 2 is heated. 

15:7. Regeneration in Refrigeration—Liquefying Gases. To liquefy 
any of the “permanent’’ gases requires that temperatures be produced 
that are far below the usual cold-room temperatures of the commercial 
refrigerating machine. For example, air at atmospheric pressure begins 
to liquefy at a temperature of about 148°R. 1 In utilizing the methods of 
refrigeration discussed to this point to produce a temperature of this 
order, a refrigerant must be selected which does not solidify at a higher 

1 The condensation of air, a mixture of gases, at constant pressure does not take 
place at constant temperature since the separate gases of which it is composed do not 
have the same saturation temperature at that pressure. The condensation of air at 
atmospheric pressure is not completed until a temperature of approximately 142°R is 
reached. During condensation the proportion of the constituents in the liquid 
changes, reaching the standard proportions for air only as condensation is complete. 


REFRIGERATION 


377 


temperature. Air itself would, of course, be a possibility, but, even at 
high pressures, the amount of reduction of temperature that can be 
obtained by throttling expansion through the expansion valve is limited 
and could not be expected to bring the temperature down from the atmos¬ 
pheric level in a single step. Adiabatic expansion behind the piston of an 
air engine would, in theory, increase the temperature drop, but, in prac¬ 
tice, the irreversibilities that accompany such a process limit the tempera¬ 
ture drop to less than the amount required. 

Some temperature drop, based on its Joule-Thomson coefficient, can be 
obtained by throttling expansion of air, even at atmospheric levels of 



Fig. 15:7. Flow diagram for adsorption refrigeration. 


temperature, and this drop is increased as the pressure from which the 
air is throttled is increased. The lower temperatures created in this man¬ 
ner can be used to chill the high-pressure air, before its expansion, to 
temperatures below that of the atmosphere, and so, after expansion, still 
lower temperatures may be created. This procedure can be continued 
until the temperature of the compressed air has been reduced to a really 
low level before its expansion, a level such that later throttling can pro¬ 
duce a temperature low enough to result in partial condensation. This is 
regenerative cooling and is the basis of the Linde process for the liquefaction 
of gases. 

The principle on which the Linde process is based is illustrated in the 
Ts diagram of Fig. 15:8 and the flow diagram of Fig. 15:9. Air at atmos- 





























378 


BASIC ENGINEERING THERMODYNAMICS 


pheric pressure and temperature is compressed in the compressor of Fig. 
15:9 to a high pressure; this is process 1-2 of Fig. 15:8 and, as shown, 
causes a considerable rise in the temperature of the air. The next step 
is to cool the air to atmospheric temperature in the cooler shown in the 
flow diagram; this process is represented by the constant-pressure line 2-3 
on the Ts diagram. Heat can be voided to the atmosphere only as low as 
atmospheric temperature, but constant-pressure cooling continues below 
T 3 as the result of regenerative heat exchange with a stream of cool air, 
the source of which will be later explained. As a result of this regenera- 



Fig. 15:8. Ts diagram—Linde process for liquefying air. 


tive cooling, the high-pressure air is cooled to state 4. From this state 
it is possible, by throttling expansion to atmospheric pressure, to pass the 
fluid into its saturated-vapor region, as is shown in Fig. 15:8. As it 
issues from the expansion valve at state 5, the fluid is a mixture of the 
liquid and the dry vapor. On the Ts diagram, the weight of liquid per 

pound of mixture at this point is represented by the ratio | en ^| 1 ^ 

1 J length ot line a-o 

The greater density of the liquid makes it possible to separate it from the 

vapor and draw it off to storage. The rest of the mixture, a dry vapor, is 

brought into communication with the line carrying the air at high pressure 

in a counterflow heat exchanger and removes heat from the high-pressure 

air as the temperature of the vapor rises, ultimately to the temperature of 

the atmosphere. Now at atmospheric pressure and temperature, it may 








REFRIGE It A TION 


379 


reenter the compressor to retrace the process, although make-up must 
also be provided to replace the weight of liquid air which was withdrawn 
below the expansion valve. 

In practice, the maximum differential of temperature that can be main¬ 
tained by the method described above is about 100°F. The liquefaction 
of air directly from atmospheric temperature would therefore not be prac¬ 
tical, but some vapor or vapors with suitable properties as refrigerants can 
be used to cover the gap in steps, or stages. This amounts to multivapor 



Regenerative 
heat exchange 


Fig. 15:9. Flow diagram—Linde process for liquefying air. 


refrigeration and is analogous to the multivapor power cycle, as applied in 
the mercury-steam binary-vapor power plant. 

Helium liquefies at about 8°R and requires the intermediate liquefac¬ 
tion of a number of vapors, hydrogen (liquefying at 36°R) being the last 
preliminary step. By reducing the pressure acting on the liquid helium, 
temperatures of about 1.5°R may be attained; refrigeration based on the 
expansion of fluids cannot carry the temperature much below this level. 
The lower temperatures required in some scientific work are obtained by 
making use of the effect of extremely low temperatures on the magnetic 
properties of some substances. Temperatures below 0.1°R have been 
demonstrated by a method based on this phenomenon. 

15:8. The Heat Pump. The heat required to maintain a comfortable 
temperature within buildings is ordinarily supplied directly from the 












































380 


BASIC ENGINEERING THERMODYNAMICS 


combustion of fuels. Work, in the form of electrical energy, may be used 
to accomplish the same purpose by passing the electric current through 
resistance heaters; but, as we have seen, work is a much more valuable 
form of energy than is heat, and this method, although sometimes used 
because of its convenience, is expensive. Moreover, it is thermodynami¬ 
cally wasteful, for a much smaller amount of work, if used to drive a heat 
pump, will account for the delivery of the required amount of heat. 
Space heating by this method, although still in the semiexperimental 
stage of development, has received considerable attention in late years, 
and there are many installations of the type in localities where conditions 
are favorable. 

The heat pump is, in every respect except one, a refrigerating machine 
in principle. The difference lies in its purpose, which is to deliver heat at 
the higher temperature T H instead of to remove heat from the colder body 
at T c . This difference, however, requires a restatement of the coefficient 
of performance as it applies to the heat pump if the cop of the heat pump 
is to represent, as does the cop of the refrigerating machine, the ratio of 
the useful effect that is accomplished to the work required to produce that 
effect. Thus 


COpheat 


pump 


JQh 

TF7 


JQc + W c 
W c 


COPrefrigerator + 1 


(15:5) 


The ideal heat pump is based on the Carnot refrigeration cvcle for which 
Qh = Qc W c 

Tc 


T h To J(T h - Tc) 

formance of the heat pump is therefore 

COPheat pump, ideal 


The ideal, or maximum, coefficient of per- 


T h 


Th - Tc 


(15:6) 


The heat-pump efficiency , analogous to the efficiency of the refrigerating 
machine, rates the performance of the heat pump by comparing its cop 
with that of the ideal heat pump, or 

COPheat pump {'l H Tc) /i r >-r\ 

VHP = -- = COPheat pump 1 - 7p -“ (15:7) 

'-^'jpheat pump, ideal H 


The obvious low-temperature body from which heat may be “pumped” 
to the higher level of temperature to be maintained within the building is 
the surrounding atmosphere. Even in the milder climates, there is a wide 
variation between the extremes of temperature of the atmosphere during 
the heating season. The heat losses from the building, and thus the heat 
which must be supplied it at the higher temperature and the weight of 
fuel that must be burned in the conventional heating process to supply 
this heat, will vary approximately directly with the differential between 









REFRIGERATION 


381 


the temperature which is to be maintained within the heated space and 
that of the outside atmosphere. On the other hand, when the heat-pump 
method of heating is used, the cop of the heat pump will decrease as this 
spread of temperature widens and more work will be required to supply a 
given quantity of heat. This makes it evident that the best argument 
for the use of heat-pump methods in place of combustion heating will 
apply to its use in milder climates. 

The variation of temperature of the earth, even at depths only a short 
distance below the surface, is much less than that of the atmosphere 
above, and the earth temperature, during the heating season, is higher 
than that of the atmosphere. This suggests that the water in wells, or 
even the earth itself, is a much more satisfactory source of the heat to be 
“lifted” to the temperature of delivery for heating purposes. A supply 
of water in the quantity required is often not available or is expensive to 
provide, and the design of heat exchangers capable of receiving heat 
directly from the earth is difficult, the performance of such exchangers 
varying greatly with soil conditions; these are the principal obstacles in 
the path of the commercial application of the heat-pump principle to the 
heating of buildings. 

The use of the heat pump for the heating of buildings is, on the basis of 
present ratios of the cost of energy in the form of heat and as work, seldom 
economical from the standpoint of operational costs alone. The dis¬ 
crepancy in cost must be balanced by considerations based on greater 
cleanliness and convenience and by the fact that the heat pump may, by 
a suitable arrangement of valves, be operated on a refrigeration cycle dur¬ 
ing the season of summer heat and used to cool the building. 

Example 15:8. A building has heat losses of 100,000 Btu/hr when the inside 
temperature is 70°F and the temperature of the air outside is 40°F. (a) Compare 

the cost of heating this building by burning oil and by using a heat pump which draws 
heat from the outside air, if the cost of oil is 6 cents per gallon, the calorific value of 
the oil is 137,000 Btu/gal, and combustion efficiency is 72 per cent and if the over-all 
heat-pump efficiency is 40 per cent and power costs 2 cents per kilowatthour. ( b ) 
Make the same comparison when the outside temperature is 10°F. 

Solution: 

(a) Combustion heater: 

100,000 , m _ 

Gallons of oil per hr - ( 0 .72) (137,000) ~ ' 

Cost of oil per hr = (1.015) (6) = 6.1 cents per hour 

Heat pump: 

cop„P = 0.4 Th T ”f c = 0.4 ( 70 5 !° 40 ) = 7.06 

, . , 100,000 141 eft t>+ /u _ (14,150) (778) 

Power to drive heat pump = -14,150 Btu/hr - (L34 )( 33> 000)(60) 

= 4.15 kw 







382 


BASIC ENGINEERING THERMODYNAMICS 


Cost of power = (4.15) (2) = 8.3 cents per hour 
cost of heat pump 8.3 


Ratio 


cost of oil 


6.1 


= 1.36 


(6) The differential of temperature is now doubled, and the heat losses will therefore 
increase to 200,000 Btu/hr. The cost of oil heating will also double to 12.2 cents per 
hour. For the heat pump: 


cop//p = 0.4 (- 70 ”° 10 ) = 3.53 

Power to drive heat pump = 1 ' = 56,600 Btu/hr = 16.6 kw 

3.53 


Cost of power = (16.6) (2) = 33.2 cents per hour 


Ratio 


cost of heat pump 
cost of oil 


33.2 

12.2 


2.72 


The heat pump is in a relatively less favorable position than under the conditions of 
part a. 

Problems 

1. Prove that all reversible refrigerating cycles must have the same coefficient of 
performance if they operate between the same hot- and cold-body temperatures. 

2. A reversible refrigeration cycle removes heat from a cold body at a tempera¬ 
ture of 10°F. The temperature of the atmosphere is 95°F. ( a ) What is its coeffi¬ 

cient of performance? (6) What is its refrigerating efficiency? (c) If reversed and 
operated as a heat engine between the same temperatures, what would be its efficiency? 

3. The temperature to be maintained in a refrigerator is 0°F. What is the ideal 
coefficient of performance if the temperature of the atmosphere is (a) 80°F; (5) 100°F? 

4. The temperature of the atmosphere is 90°F. What is the ideal coefficient of 
performance if the temperature of the cold body is (a) 20°F; ( b ) 0°F? 

5. A refrigerating machine removes heat from a cold room at 10°F and discharges 
it to the atmosphere at 100°F. Its refrigerating efficiency is 55 per cent. What 
power is required per ton of refrigeration? 

6. A refrigerating machine with a refrigerating efficiency of 50 per cent discharges 
heat to the atmosphere at 90°F. Its coefficient of performance is 3.5., What is the 
temperature of the cold body? What horsepower is required per ton of refrigeration? 

7. Assume that the plant diagramed in Fig. 15:3 is charged with a gas such as 
air, instead of a vapor. If used to remove heat from a cold room at Tcand discharge 
it to the atmosphere at Th, what is the highest temperature that can be attained by 
the gas as it receives heat in passing through the cold room? What is the lowest 
temperature which the gas may reach as it delivers heat to the atmosphere? Assume 
that the compression and expansion processes are isentropic, and draw the cycle on 
Ts coordinates, comparing it with a Carnot cycle which operates between the same 
temperatures of cold room and atmosphere. In what sense is the gas cycle you have 
plotted a reversible cycle, and under what conditions is it not reversible? Discuss 
the relative advantages of a gas and a vapor as refrigerating mediums, based upon 
the Ts diagram. 

8. In the gas refrigeration cycle that has been plotted on a 7s diagram in Prob. 7, 
assume the gas to be air, the temperature and pressure at the beginning of compression 
being 0°F and 30 psia. The temperature at the beginning of expansion is 90°F; the 
pressure is 70 psia. The heat-exchange processes are assumed to take place at con¬ 
stant pressure, and both compression and expansion are isentropic. (a) What is the 








REFRIGERA TION 


383 


pressure of the air in the heat exchanger between compressor and engine? (6) Calcu¬ 
late the heat removed from the cold body, the compressor work, engine work, and 
the net work of the cycle per pound of air flow around the circuit, (c) Calculate the 
cop of the cycle, and compare with that computed in Example 15:2, when ammonia 
was used as the refrigerant. ( d ) Calculate, per ton of refrigeration, the mass flow 
rate, the volume of air per minute entering the compressor, and the net power required. 
Compare with the similar results in Example 15:2. (e) What is the refrigerating 

efficiency of the air cycle, based upon constant temperatures of cold room and 
atmosphere? 

9. Work Example 15:2 using (a) sulfur dioxide, ( b) F-12 Freon as the refrigerants. 
Compare with the results of the example, and discuss. From the standpoint of its 
ideal cop, which of the three vapors is the best? Could water vapor be used as the 
refrigerant under the conditions of the problem? 

10. Discuss the effect of charging the plant that is diagramed in Fig. 15:5 with a 
perfect gas instead of a vapor. Show that in that case a temperature below that of 
the atmosphere could not be maintained in the cold room. 

11. Work Example 15:3 for dry compression, using sulfur dioxide as the refrigerant. 
Compare with the results of the example, and discuss. From the standpoint of the 
cop of their cycles, which of the two vapors has the advantage under the conditions 
as stated? Contrast with the similar comparison that was made in Prob. 9. What 
difficulties would be encountered in the cycle if H 2 0 were used as the refrigerant? 

12. Work Example 15:3 for wet compression, using (a) sulfur dioxide, (6) F-12 Freon 
as the refrigerants. Compare the coefficients of performance of the three refrigerants 
(include ammonia in the comparison) under the conditions of the example, and 
discuss. 

13. Based on a cold-room temperature of 40°F and atmospheric temperature of 
80°F, calculate the wet-compression coefficients of performance of the vapor-com¬ 
pression refrigeration cycle with (a) ammonia, ( b ) sulfur dioxide, (c) carbon dioxide, 
(d) F-12 Freon, and (e) H 2 0 as the refrigerants and compare. Compare the volumes 
that must be handled by the compressor per ton of refrigeration and the pressure 
range of compression. 

14. Based on a cold-room temperature of 40°F and atmospheric temperature of 
80°F, calculate the dry-compression coefficients of performance of the vapor-com¬ 
pression refrigerating cycle with (a) ammonia, (6) sulfur dioxide, and (c) H 2 0 as the 
refrigerants, and compare. Per ton of refrigeration, compare the volumes that 
must be handled by the compressor. What is the pressure range of compression in 
each case? 

15. In Example 15:3, if the adiabatic efficiency of the compressor is 80 per cent in 
the dry-compression cycle, at what temperature does the ammonia leave the com¬ 
pressor? What is the coefficient of performance of the resulting cycle, assuming 
that points 1, 2, and 3 of the cycle are the same as in the example? What is its 
refrigerating efficiency? 

16. Leaving the absorber of Fig. 15:6, the aqua ammonia has a temperature of 
80°F, a pressure of 38 psia, a weight concentration x of 0.29, and a specific enthalpy 
of —40 Btu/lb. Aqua ammonia is withdrawn from the generator at a temperature 
of 305°F, a pressure of 180.6 psia, a concentration of 0.13, and a specific enthalpy of 
239 Btu/lb. Assuming complete rectification, (a) what weight of strong solution 
must the pump handle per pound of ammonia passing through the expansion valve? 
(5) What weight of weak solution is returned to the absorber from the generator? 
(c) What minimum work is required to drive the pump if the aqua ammonia is con¬ 
sidered incompressible and to have a density of 60 lb /ft 3 ? 


384 


BASIC ENGINEERING THERMODYNAMICS 


17. (a) Based on the data and results of Prob. 16 and assuming that heat is not 

exchanged between the weak and the strong solutions and that the ammonia enters 
the absorber from the evaporator as a saturated vapor at 10°F with a specific enthalpy 
of 537 Btu/lb, 1 calculate the heat that must be removed from the absorber per pound 
of ammonia that passes through the expansion valve ( Q a ). ( b ) If a heat exchanger 

is employed and, per pound of ammonia passing through the expansion valve, 950 Btu 
of heat leaves the weak solution and enters the strong solution in the course of their 
flows between absorber and generator, what is the specific enthalpy of the weak 
solution as it enters the absorber and the specific enthalpy of the strong solution as it 
enters the generator? What is Q a per pound of ammonia passing the expansion 
valve? 

18. Continuing the study of the ammonia-absorption plant of Probs. 16 and 17 
and making use of the data and results of those problems, assume that the vapors 
leaving the generator enter the rectifier at 305°F, 180.6 psia, and that the weight 
concentration of ammonia in these vapors is 0.60 and their specific enthalpy is 890 Btu 
per pound of mixture. Returning to the generator from the rectifier is aqua ammonia 
at 305°F, 180.6 psia, an ammonia weight concentration of 0.13, and a specific enthalpy 
of 239 Btu/lb. The pure ammonia leaving the rectifier and entering the condenser 
is a saturated vapor at 90°F. Per pound of ammonia passing from rectifier to con¬ 
denser, find (a) the weight of vapor mixture entering and (6) the weight of aqua 
ammonia leaving the rectifier, (c) Calculate the heat (Q g ) that must be supplied 
to the generator per pound of pure ammonia passing the expansion valve. Assume 
that the heat exchanger of Prob. 17, part b, is employed. ( d ) Calculate the heat 
that must be removed in the rectifier ( Q r ) per pound of ammonia passing the expan¬ 
sion valve, (e) Assuming that the pure ammonia enters the expansion valve as a 
saturated liquid, calculate Qh and Qc per pound of ammonia passing that valve. 
(/) Prepare an energy-balance for the entire plant, based on the First Law. 

19. Calculate the coefficient of performance for the ammonia-absorption refriger¬ 
ation plant of Probs. 16 to 19. What is the lowest temperature that may be substi¬ 
tuted for Ts? What is the highest temperature that may be used as Th ? What is 
the refrigerating efficiency of the cycle? 

20. Equation (15:4) is an approximation that ignores the work of the liquid pump 
of the absorption-refrigeration cycle. Based on including that work, what coefficient 
of performance would be obtained in Prob. 19? 

21. A building has heat losses that total 150,000 Btu/hr when the inside tempera¬ 
ture is 70°F and the outside temperature is 30°F. If a heat pump is used to heat 
the building and it has an efficiency of 45 per cent, what horsepower will be required 
when the temperatures are as specified above? What is the coefficient of performance 
of the heat pump? What are the similar answers if the outside temperature drops 
to 20°F? 

22. When the inside temperature is 70°F and the outside temperature is 35°F, 
a building has heat losses of / 5,000 Btu/hr. The coefficient of performance of a 
heat pump which draws heat from the atmosphere and is used to heat this building 
is 6.5. When the temperatures are as specified above, what power is required? At 
what rate is heat removed from the atmosphere? What is the efficiency of the heat 
pump? 

• 

This enthalpy may be obtained from the Bureau of Standards table of the proper¬ 
ties of ammonia by subtracting 78 Btu from the values shown in that table. This is 
done in order that they will be based on the same reference level as the Jennings and 
Shannon tables, from which the enthalpy of the weak and the strong solutions has 
been taken. 


REFRIGERA TION 


385 


Symbols 

cop coefficient of performance 
h specific enthalpy 
J proportionality factor 
M mass rate of flow 
p pressure, psi 

P pressure, psf; pressure in general 
Q rate of heat flow 
s specific entropy 
T absolute temperature 
v specific volume 
V volume; volume rate of flow 
W rate of work delivery 

x quality of the vapor; also, concentration of ammonia in aqua ammonia 
Greek Letters 

7j r refrigerating efficiency 
V HP efficiency of a heat pump 

Subscripts 

a absorber 
c compressor 
C cold body 
e engine 

/ of the saturated liquid 

g of the saturated vapor; also, generator of the absorption cycle 
H hot body 
HP heat pump 
l lower 
R refrigerator 
S source 
u upper 


CHAPTER 16 


THE EVALUATION OF IRREVERSIBILITY 

16:1. Introduction. A heat engine operates on a cycle with the pur¬ 
pose of accomplishing a continuous conversion of energy from the form of 
heat into the form of work. Its effectiveness in fulfilling that purpose is 
measured in terms of the ratio of the rate at which work is delivered to 
that at which heat is received; no heat engine can convert all of the heat 
that it receives into work. Its performance is rated by comparing it with 
that of the ideal reversible heat engine which receives heat under the 
same conditions and exhausts it to the same reservoir (refrigerator, or 
sink), and this reservoir is characteristically represented, for the engineer, 
by the atmosphere. Any failure to attain the ideal performance associ¬ 
ated with this reversible engine can be attributed to irreversibilities that 
have been encountered in carrying out one or more of the processes of the 
heat-engine cycle. These processes are ordinarily carried out in the real 
power plant in separate devices, each responsible for the demonstration 
of a part of the entire cycle. In order to improve the performance of the 
entire heat engine, by reducing its irreversibility, it is desirable that each 
of the processes making up the cycle be individually studied for its irre¬ 
versibilities; in that way, the source, or sources, of the trouble may be 
located and may receive proper attention. Any other method may be 
misleading. For example, the steam, as it enters the condenser of the 
steam power plant, contains large amounts of energy which are immedi¬ 
ately discarded to the atmosphere; this is sometimes listed as the “exhaust 
loss” in analyzing the performance of the engine. But even the ideal 
Carnot engine has a large “exhaust loss,” which is based on the standards 
set by the Second Law, and attention given to reducing this “loss” will 
be rewarding only if it is larger than is warranted under the conditions 
that surround the operation of the engine and if the source of the irre¬ 
versibility which accounts for the excess is known. 

A similar discussion will apply to the execution of a cycle which has 
for its purpose the maintenance of a temperature below that of the atmos¬ 
phere within a limited space, as by a refrigerating machine, or the delivery 
of heat taken from the atmosphere to a body at higher temperature, the 
purpose of the heat pump. The performance of plants of this kind can be 
improved only within the limits of performance set by the ideal (reversi¬ 
ble) refrigeration cycle, and this improvement is possible only as the 

386 


THE EVALUATION OF IRREVERSIBILITY 


387 


irreversibilities encountered in the individual processes of the cycle are 
reduced. Unless and until the magnitude of the irreversibility encoun¬ 
tered in each individual process of the real cycle is calculated and con¬ 
sidered as to its effect on the performance of the entire machine, effort 
may not be intelligently directed toward improvement of that performance. 

16: 2. Availability. A system, as it enters a process at a state which 
will be designated by the subscript 1, possesses a store of energy dependent 
on its pressure, its temperature, its position, its motion, and its electricity 
and magnetism. In combination with a vast medium , or atmosphere, 
having the pressure P 0 and the temperature T 0 , it will be capable of deliv¬ 
ering a portion of this store of energy in the form of work to external sys¬ 
tems other than the medium as it shifts to a dead state , at which it is at 
rest within the medium and is in equilibrium with it as regards pressure, 
temperature, and electricity and magnetism. On the other hand, if it is 
already at the dead state, no work may be obtained from the system or 
from the medium either individually or as the result of interaction between 
the two. 1 

If the system is not at its dead state, a tendency will exist for a change 
toward the dead state. In order to initiate this change of state, it may be 
necessary for some external system (other than the medium) to furnish the 
necessary disturbance, but, once started, a portion of the work released 
as a result of the change of state may be used to restore to this external 
system the energy that has been borrowed from it, and, because of the 
spontaneous character of the change, it will always be possible to deliver 
an excess of (positive) work to the surroundings. This will be true even 
if, originally, the temperature and/or the pressure are below those of the 
medium. For if the temperature of the system is originally less than that 
of the medium, heat may be supplied by the medium to a Carnot engine 
which has the system as its sink and work will be obtained while the 
temperature of the system, as it receives heat from the exhaust of this 
Carnot engine, gradually rises toward the dead-state temperature; simi- 

1 It is assumed that there is no relative movement between parts of the medium, 
nor does any difference in pressure (at the same level), temperature, or composition 
exist throughout its extent; otherwise, the medium alone would always constitute a 
source of work. When, in the course of a process, the system combines, either chemi¬ 
cally or mechanically, with a portion of the medium, as in the combustion of a fuel, 
that portion is made a part of the system, the latter being described to include it 
throughout the entire process. Thus, in the combustion process, the system is 
described to include, at the beginning of the process, the fuel plus that portion of the 
atmosphere which will ultimately enter into the process. The medium does not 
include this portion, which, in the course of the process, may change in temperature or 
pressure; the medium is always distinguished by a constant pressure at any given 
level and by a constant temperature. Moreover, because of its vast extent, the 
medium may receive or give up energy in the form of heat or work without any change 
in either temperature or pressure. 


388 


BASIC ENGINEERING THERMODYNAMICS 


larly, when Pi < P 0 , this differential of pressure may be used to do work 
on a piston as the pressure of the system gradually rises to the dead-state 
pressure. 

We may therefore say that, for any state of the system, the maximum 
work that can be produced by the system and the medium, as the former 
changes to its dead state, cannot be less than zero. This maximum work, 
when obtained without the aid of other than cyclic changes on the part of 
external systems other than the medium, is called the availability which 
corresponds to that state; the availability which corresponds to the dead 
state is zero, for all other states must be positive. 

It has been shown (Art. 5:3) that maximum work is always associated 
with the reversible process. During any change in state of the system 
toward its dead state, the availability will always decrease. If a reversi¬ 
ble process that connects the initial and final states is applied, the maxi¬ 
mum amount of work which might have been released to external systems 
other than the medium as the result of interaction between the system 
and the medium during this change in state may be calculated; that 
amount of work will represent the decrease in availability of the system 
(in combination with that medium) as a result of the change in state. 
Conversely, if the change is away from the dead state and therefore 
toward greater availability, it is required that energy must have entered 
the system from external systems other than the medium. If this change 
in state is reversibly accomplished, the amount of work which the system 
has constructively received in the course of the process from external sys¬ 
tems other than the medium will be a minimum ( i.e ., with respect to the 
system, a positive-work maximum) and will represent the increase of 
availability that has taken place. The word “ constructively, ” as used 
in the foregoing sentence, means that that portion of the heat received by 
the system during the process which could have been converted into work 
if supplied to a Carnot engine that has either the system or the medium 
as its source (depending upon which is at the higher temperature) and 
the other as its refrigerator must be counted as work. The constructive 
work of a reversible process that connects states of equal availability is, 
of course, zero. 

In order to calculate the change in availability that accompanies a 
change in state, it will be necessary to evaluate the maximum amount of 
work that can be developed by interaction between the system and the 
medium as the system changes from state 1 at the beginning of a given 
process to state 2 at the end of that process; this, as stated above, is 
determined by assuming the change in state to have taken place reversi¬ 
bly. During an infinitesimal section of this assumed reversible process 
there may be an infinitesimal heat flow dQ and an infinitesimal flow of 
energy in the form of work, dW. Both dQ and dW are expressed, in sign, 


THE EVALUATION OF IRREVERSIBILITY 


389 


with reference to the system. If, during this infinitesimal step, the tem¬ 
perature of the system is below that of the medium, the sign of dQ is posi¬ 
tive; dQ is negative in sign if the temperature of the system exceeds that 
of the medium. If the pressure of the system is greater than that of the 
medium, dW is positive, and vice versa. 

Let us first consider the work that may be delivered to external systems 
by reason of a heat flow between system and medium. If the temperature 
T of the system is above that of the medium during this infinitesimal part 
of the entire process and if this heat dQ had been supplied a Carnot engine, 
the work delivery of that engine would have been -dQ(T - T 0 )/T, the 
minus sign having been given dQ since the direction of heat flow with 
respect to the engine is opposite to its direction with respect to the system. 
If, on the other hand, T is less than T 0 , the heat supplied the Carnot 
engine will come from the medium and will be equal to dQ(T 0 /T), where 
dQ is still the heat flow with respect to the system. The work output of 


the Carnot engine will, in that case, be dQ 


To T 0 - T 


T ( 



thus being represented by the same expression that resulted from the 
opposite assumption. Moreover, this expression must give a positive 
result, indicating that work was delivered to external systems, whenever 
T differs from T 0 ; for if T > T 0 , dQ is negative, and if T < T 0 , dQ is 
positive. 

To obtain the total work deliverable to external systems, account must 
also be taken of dW, the work flow during this infinitesimal part of the 
reversible process. This d W is, of course, the gross work of the process 
and not the net work deliverable to external systems other than the 
medium; it is the net work in which we are interested. If the pressure P 
of the system is above P 0 , the system will expand and the work done by 
the system on the medium as the result of this expansion is P 0 dV; if 
P < Po, dV is negative, the work done by the system on the medium is 
negative but, since dV is negative, still may be expressed as P 0 dV. We 
may therefore write for the net work deliverable to external systems, 


dW - Po dV. 

Adding the work deliverable by reason of a pressure difference between 
system and medium to that obtainable because of a temperature differ¬ 
ence, we have 


^T -To , aw Po dV 

dQ -~l— + T ~ VT 



Po dV 
J 

Po dV , ^ dQ 
J + lo T 


But, from Eq. (2:4), — dQ + dW/J = —dE ; also, since this is a reversi¬ 
ble process, we may substitute dS for dQ/T. Thus the maximum work 











390 


BASIC ENGINEERING THERMODYNAMICS 


deliverable to external systems during this infinitesimal process is 

— dE — + TodS 

Since the pressure P 0 and the temperature T 0 of the medium do not 
change, this may be extended to cover a finite change of state, as follows, 

-A E - + T„ AS 

and the maximum possible amount of work that can be delivered to 
external systems is therefore the decrease in the quantity 

E + ~f~ToS (16:1) 


Because this maximum amount of work measures the availability corre¬ 
sponding to a given state of the system, the foregoing expression will be 
called the availability function. It is not a property of the system alone, 
being expressed in terms of system properties and the temperature and 
pressure of the medium. 

The availability of a system at a given state may be measured as the 
change in the value of the availability function as the system changes 
from that state to its dead state. At this dead state, the properties of 
the system will be designated by the subscript 0, and we may write 


Availability 





P oV 0 
J 



(16:2) 


This expression cannot, for any state of the system, have a lower value 
than zero. 1 The change of availability as the system changes from state 
1 to state 2 is the change in the value of the availability function between 
the two states, or 


A (Availability) = 





(16:3) 


This change in availability may be either positive, indicating an increase 
in availability, or negative, if the system, in combination with the medium, 
is less capable of delivering work to external systems as it changes to the 
dead state. 

When the system is a simple system, energy storage by reason of motion, 
elevation, magnetism, or electricity being negligible, U may, of course, 
replace E in the foregoing expressions. For the simple system, the sym- 

1 The availability function may have a value less than zero, but its value at the 
dead state is its lowest possible value. 









THE EVALUATION OF IRREVERSIBILITY 


391 


bol (3 will be used to designate the value of the availability function as it 
applies to a unit weight, or 


0 = u -f 


Pov 

J 


T 0 s 


(16:4) 


Example 16:2. In combination with an atmosphere at 14 psia, 70°F, calculate 
the availability of 1 lb of H 2 0 at a pressure of 200 psia and in the following states: 
(a) 500°F; (6) saturated vapor; (c) saturated liquid; (d) 70°F; (e) 32°F. 

Solution. The /3 function will represent the availability function for this simple 
system. In the dead state, the water has a pressure and temperature corresponding 
to those of the atmosphere. It is thus a compressed liquid but, for simplicity of 
calculation, will be assumed incompressible. This means that its volume, internal 
energy, and entropy are the same as those of the saturated liquid at the same tem¬ 
perature. Thus, at p = 14.7 psia, t = 70°F, 


~ _ 38.04 - ( 14 ^H 14 - 7 - 0-86) (0.01606) _ 3§ 0Q Btu 

J 7/o 

y 0 = 0.01606; So = 0.0745 

/3 0 = w 0 + ~ Toso = 38.00 + ( 144 )( 14 JH°- 01606 > _ (530) (0.0745) = 38.00 


J -- 1 778 

(a) This is a superheated steam condition. 


+ 0.04 - 39.49 = -1.45 Btu 


u = 


0 = 


h ^ = 1268.9 - _ „ 68 . 0Btu 

1168.0 + (144)(1 ll g )(2 '' 26) - (530)(1.6240) = 1168.0 + 7.4 

/ / o 


860.7 

= 314.7 Btu 


Availability = /3 — /3 0 = 314.7 — ( — 1.45) = 316.1 Btu 


0 b ) /3 = 1113.7 + 


(144) (14.7) (2.288) 
778 


- (530) (1.5453) = 300.9 Btu 


Availability = 300.9 — ( — 1.45) = 302.3 Btu 
(c) p = 354.68 + (144)(14 _ 7 >(°- 01839) _ (530)(0.5435) = 66.68 Btu 


Availability = 66.68 — ( — 1.45) = 68.13 Btu 

(d) u = uo = 38.00; v = v 0 = 0.01606; s = s 0 = 0.0745 
(3 = 0 O = —1.45 Btu; availability = 0 

This result is slightly misleading, though correct as based on the assumption of 
incompressibility. If compressibility had been taken into account, /3 would have been 
very slightly larger (less negative) than (So and the availability would have been 
positive. The number of places to which the values of the properties are shown in 
the steam tables would hardly have been sufficient to show this difference. 

(e) ft = 0.00 + (144)(14 - 7 7 ) r (0,016 ° - - (530) (0.00) = 0.04 Btu 












392 


BASIC ENGINEERING THERMODYNAMICS 


Availability = 0.04 — ( — 1.45) = 1.49 Btu 

Note that positive availability is shown although the temperature of the system 
is below the temperature of the atmosphere. 


16 : 3. The Evaluation of Irreversibility. There have been occasions in 
earlier chapters when we were concerned with the evaluation of irreversi¬ 
bility. For example, comparisons have been made between the efficiency 
of a heat engine, or a heat-engine cycle, and the efficiency of its equivalent 
Carnot cycle; the refrigerating efficiency introduced in Chap. 15 is 
another example. Both of these developed as the ratio of realized to per¬ 
fect performance and applied to the cycle as a whole. Engine (or turbine) 
efficiency was, on the other hand, an attempt to express the irreversibility 
of one of the processes of the cycle in the form of a similar ratio. 

We shall now introduce a method of evaluating the irreversibility of a 
process in terms of the difference between the amount of useful work 
which might have been delivered as the system changed between the end 
states of the process if that process had been reversibly performed and 
the work that was actually (though sometimes, at least in part, con¬ 
structively) delivered during the real process. The maximum work that 
is associated with the reversible process has been shown to be equal to the 
decrease of availability; for an infinitesimal segment of the entire process, 
it is — dE — P 0 dV/J -\- T 0 dS. In calculating the constructive work of 
the real process, we may first set down the work actually delivered as 
useful work performed on external systems other than the medium. This 
is dW — Pq dV and will be designated as dW u . To this useful work must 
be added the work constructively received by these external systems due 
to the amount of heat they have received; if this heat dQ had been sup¬ 
plied a Carnot engine having the medium as its sink, the work realized 
would have been —dQ(T - T 0 )/T, where T is the temperature of the 
system as this minute step is taken. Note that it has been shown in Art. 
16:2 that, if the temperature T is below T 0 , this expression will still be 
valid. The total constructive work of the real, though infinitesimal, 
process is therefore dW u /J - dQ(T - T 0 )/T, and we may write 

d(Irreversibility) 

- (-<« - 1 + Tods) - 7 Tjr'j 

+ <«*> 

The irreversibility encountered in the course of the entire process may be 
evaluated by integrating this expression, or 


Irreversibility = 



~ A (availability) 


(16:6) 










THE EVALUATION OF IRREVERSIBILITY 


393 


An expression for irreversibility that is sometimes easier to use than Eq. 
(16:6) may be obtained from that equation by writing it in the form 


Irreversibility 



1Q2 


E‘ 


dQ 


- £1 + Po(F ^ Fl) _ To(S 2 - Si) 

- ( £, _ Ei) - + p »(v* - y.) 

J 


+ T 0 (S 2 - SO 


Or, since iW u2 + P 0 (V 2 — Vi) = iW 2 and 1 Q 2 = E 2 - E x + AV 2 /J, 

Irreversibility = 7 7 0 (*S 2 — Si) — T 0 (16:7) 

Equations (16:6) and (16:7) will both find use in measuring the irreversi¬ 
bility of the closed system during a nonflow process. The irreversibility 
can never be negative as that would imply that the process had been more 
perfectly performed than a reversible process which connects the same 
end states. 

Example 16:3. The temperature and pressure of the atmosphere are 70°F and 
14.7 psia, respectively. The system consists of 1 lb of steam, originally at 200 psia, 
500°F. Calculate the change of availability and the irreversibility which accompany 
the following processes: (a) The system, while confined behind a piston at constant 
pressure, expands slowly as its temperature is increased to 600°F by the addition of 
heat. ( 6 ) The same change of state as in part a is effected without heat flow, as the 
result of a paddle-wheel process, (c) The system expands adiabatically and reversi¬ 
bly to a pressure of 14.7 psia. (d) It expands adiabatically into an adjoining space 
which is initially empty. The final pressure is 14.7 psia. (e) It expands adiabatically 
to become a saturated vapor at 14.7 psia. 

Solution. The P function becomes the availability function for this simple system. 
pi has been calculated as 314.7 Btu in Example 16:2, part a. 

(a) M2 _ , 322 .! _ (144) (200^( 3^06 0) . lmg Btu 

/3 2 = 1208.8 + ( 144 ) (14^/) (3.06°) _ ( 530 )( 1-6767 ) = 328.4 Btu 

7 7 o 

A (Availability) = /3 2 — Pi = 328.4 — 314.7 = 13.7 Btu 
In Eq. (16:7), 



Irreversibility = T , 0 (s 2 — Si) — To(s 2 — Si) = 0 

(b ) A(Availability) = 13.7 Btu from part a 
In Eq. (16:7), 












394 


BASIC ENGINEERING THERMODYNAMICS 


dQ = 0 

Irreversibility = To(s 2 — si) — 0 = 530(1.6767 — 1.6240) = 28.0 Btu 

This means that the net work of stirring was 28.0 + 13.7 = 41.7 Btu. 
Checking, 

(P — Po)(v 2 — Vi) 


Net work = h 2 — hi — 


J 


= 1322.1 - 1268.9 - 144(200 - 14.7)(3.060 - 2.726) _ 4 , ? Btu 

/ i o 

(c) Using the Mollier diagram, the final condition is located at x 2 = 0.908; u 2 = 
180.0 + (0.908)(897.5) = 995.0 Btu; v 2 = (0.908)(26.80) = 24.34; s 2 = s 1 = 1.6240. 

0 2 = 995.0 + ( 14 4)(!4 7)(2 _ 4 . 34 ) _ (530)(1 624 o) = 200.5 Btu 

/ / O 

A (Availability) = 0 2 — 0i = 200.5 — 314.7 = —114.2 Btu 

In Eq. (16:7), 

dQ = 0 and s 2 — Si = 0 
Irreversibility = 0 

This means that 114.2 Btu of useful work resulted from the expansion. 

Checking, 


iWt 

J 

iWut 

J 


= Ui - u 2 = 1168.0 - 995.0 = 173 Btu 

_ iW 2 P 0 (v 2 - Vl ) = 1730 _ (144)(14.7)(24,34 - 2.726) = m 2 Btu 


J 


J 


778 


{d ) There is no work and no heat flow. Therefore, for this nonflow expansion, 
u 2 = ui = 1168 Btu. Using information from Table 3 of the steam tables, the final 
state is located as corresponding to a temperature t 2 of approximately 463°F. Also, 
v 2 = 37.27, and s 2 = 1.9075. Then 

0 2 = 1168.0 + ( 144 )( 14 - 7 )( 3 ^- 27 ) _ ( 530 ) (1.9075) = 258.3 Btu 

/ / O 

A (Availability) = 02 — 0i = 258.3 — 314.7 = —56.4 Btu 
Using Eq. (16:7), 

Irreversibility = T 0 (s 2 — Si) — 0 = 530(1.9075 -*• 1.6240) = 150.1 Btu 

The difference between the irreversibility and the decrease of availability is 150.1 
— 56.4 = 93.7 Btu. Checking, it will be found that this was the amount of work 
required to evacuate the atmosphere from the space into which the expansion took 
place and thus was required from external systems preparatory to the execution of the 
designated process. 

(144) (14.7) (26.80) 


(e) 02 = 1077.5 + 


778 


(530) (1.7566) = 219.4 Btu 


A(Availability) = 0 2 — 0 1 = 219.4 — 314.7 = —95.3 Btu 
From Eq. (16:7), 

Irreversibility = T 0 (s 2 — sQ — 0 = 530(1.7566 — 1.6240) = 70.3 Btu 
For this adiabatic process, the difference between the decrease of availability and the 
irreversibility should equal the useful work. 












THE EVALUATION OF IRREVERSIBILITY 


395 


Checking, 

y 2 = Mi - uj = 1168.0 - 1077.5 = 90.5 

'N-' = l W t - P >0»-ri = 90.5 - (lM)(,liZ ) . (26^0 - 2.726) _ ^ 

J J i (8 

= 95.3 - 70.3 


16 : 4. Availability in Steady Flow. Most real engineering processes are 
essentially steady-flow processes. The availability in steady flow 
includes the work which can be delivered to external systems other than 
the medium by reason of that flow. This increment of useful work is the 
difference between the work required to displace the fluid across the sec¬ 
tion at which the availability is to be measured (the flow work of Chap. 
3) and the amount of that work if the volume displaced had been the same 
but the pressure had been that of the medium; it is thus equal, per pound 
of fluid flowing, to Pv — P 0 v. Returning to Eq. (16:2) and making this 
addition to the availability of a unit weight, we have 

( | P 0^0 rp 

— I Co H— j- — 1 

Pv — PqV 
+ J 

= (e + y - T„s) - (e„ + (16:8) 


Availability per lb = ( e + 


PqV 

J 


Tos 


If energy storage due to capillarity, magnetism, or electricity is negligible, 
e = u + V 2 /2Jg + z/J and 


/ Pv V 2 z\ 

Availability per lb = ( u + -j — T 0 s + + -jj 

( | PqVq y, , To 2 ■ Zo 

-\ Ua + —~ ToSo + 2Jg + J 
But u + Pv/J = h and V 0 is zero (see footnote, page 387). Therefore 


Availability per lb = 


Zo 


h — T qS + %Tg J/ ~~ ^ _ ToSo -b 

b + ir g + 7) ~ ( bo + 7) (16:9) 


where 


b = h - T 0 s (16:10) 


Like the availability function developed in Art. 16:2, b is a combination 
function of the properties of the system and the medium. It cannot, 

* Note that the availability in steady flow may be negative if the pressure of the 
substance is less than that of the medium. 















396 


BASIC ENGINEERING THERMODYNAMICS 


however, be called a steady-flow availability function, since the latter 
must include the energy quantities that are expressed in terms of the 
velocity and the elevation, or 


Availability function, steady flow = b + 


V 2 

2Jg 



(16:11) 


Thus the change in availability in steady flow, as the flow progresses from 
section 1 to section 2, is 


A (Availability, steady flow) 



VV 
2 Jg 





(16:12) 


16: 5. Irreversibility in Steady Flow. In evaluating the irreversibility 
of a steady-flow process, the basis of evaluation is the same as for the 
closed-system process, the work constructively delivered to external sys¬ 
tems in the course of the process being deducted from the decrease of 
availability that has resulted. Thus 


Irreversibility, steady flow 



— A (availability, steady flow) (16:13) 


where \W 2 has the same meaning as in the steady-flow energy equation 
[Eq. (3:5)] and is the work delivered to external systems other than the 
medium as the result of the real steady-flow process. Both iTE 2 and dQ in 
this equation are expressed in sign with respect to the system, so that a flow 
of work from the system is positive, a flow of heat negative. Although 
the steady-flow availability may conceivably be negative, irreversibility 
in steady flow can never be less than zero. 

Example 16:5A. A steady stream of H 2 0 at 200 psia, 500°F, enters the following 
steady-flow processes at negligible velocity. The pressure and temperature of the 
atmosphere are 14.7 psia and 70°F. Calculate, for each process, the change of avail¬ 
ability and the irreversibility, per pound of H 2 0. (a) The stream is heated slowly 

at constant pressure until its temperature is 600°F. ( b) It expands adiabatically and 

reversibly through a nozzle to 14.7 psia. (c) It is throttled to 14.7 psia, the final 
velocity being negligible. 

Solution: 

hr = 1268.9; s x = 1.6240; hr = 1268.9 - (530)(1.6240) = 408.2 Btu 
(a) h 2 = 1322.1; s 2 = 1.6767; h 2 = 1322.1 - (530)(1.6767) = 433.4 Btu. 

A(Availability) = b 2 — bi = 433.4 — 408.2 = 25.2 Btu 

/ 2 nr* rn 

“ — J ,—- 0 dQ, when 

evaluated, will equal 25.2 Btu. This may be checked, approximately, as follows: 








THE EVALUATION OF IRREVERSIBILITY 


397 


The system receives 1322.1 — 1268.9 Btu of heat at an average temperature of 550°F. 
If this amount of heat were supplied at that temperature to a Carnot engine with a 
70°F sink, the work output of that engine would be 53.2(550 - 70)/(550 + 460) = 
25.3 Btu. Thus the work constructively furnished by external systems accounts for 
the increase in availability, confirming the reversibility of the process. 

(6) From Example 16:3, x 2 = 0.908. h 2 = 180.07 + (0.908)(970.3) = 1061.1 Btu; 
S 2 = Si = 1.6240; b 2 = 1061.1 — (530)(1.6240) = 200.4 Btu. The availability func¬ 
tion as expressed for the nozzle exit must also include the kinetic energy term V 2 2 /2Jg 
= h x - h 2 = 1268.9 - 1061.1 = 207.8 Btu. Then 

Vo 2 

A (Availability) = b 2 + ~ - b i = 200.4 + 207.8 - 408.2 = 0 

This value is in accordance with the fact that there has been no exchange of energy 
with external systems either as heat or as work and with the description of the process 
as reversible. 

(c) hi — 1268.9 = h 2 . At 14.7 psia, this enthalpy corresponds to a temperature of 
approximately 462°F and s 2 = 1.9070; b 2 = 1268.9 - (530)(1.9070) = 258.2 Btu. 

A (Availability) = b 2 — bi = 258.2 — 408.2 = —150.0 Btu 

There has been no heat flow or useful work. Applying Eq. (16:13), the irreversibility 
is therefore the decrease of availability, or +150.0 Btu. 

It is remembered that the various items of equipment in an engineering 
plant usually operate to accomplish certain changes in the condition of the 
fluid while the latter is in essentially steady flow. Equation (16:13) may 
then be applied to each unit in turn, and each unit may thereby be 
charged with its proper portion of the blame for any failure of the entire 
plant to show a perfect performance of its function. In large proportion, 
the various pieces of apparatus that operate on the fluid in the engineering 
plant may be grouped as belonging either to the engine classification or in 
the category of heat exchangers. 

The class of apparatus to which the engine belongs (and this includes 
turbines and compressors) carries out a process which is essentially 
adiabatic, even though not reversible, in character, and dQ in Eq. (16:13) 
is therefore zero. This simplifies the calculation of the irreversibility 
encountered in the type of process carried out by this group to a simple 
subtraction of the useful work delivered from the decrease of availability. 
When the unit is a compressor, an increase of availability is usually 
detected but in this case the work is negative, so that the irreversibility 
will be characteristically positive as for the engine. 

Example 16:55. The temperature and pressure of the atmosphere are 70°F and 
14.7 psia. Steam at 200 psia, 500°F, enters a turbine at low velocity and is expanded 
to 14.7 psia. The turbine has an efficiency of 70 per cent, and the exit velocity is 
negligible. Calculate the change of availability per pound of steam and the irreversi¬ 
bility of the turbine process. 

Solution: 


hi = 1268.9; si = 1.6240; 5, = 408.2 Btu 


[Ex. 16:5A] 


398 


BASIC ENGINEERING THERMODYNAMICS 


At the end of an isentropic expansion to 14.7 psia, the enthalpy is 1061.1 Btu. Then 


h 2 = 1268.9 - 0.70(1268.9 - 1061.1) = 1123.4 Btu 
1123.4 - 180.07 


x 2 = 


= 0.972; s 2 = 0.3120 + (0.972)(1.4446) = 1.716 


970.3 

b 2 = 1123.4 - (530)(1.716) = 213.9 Btu 
A (Availability) = b 2 - b x = 213.9 - 408.2 = -194.3 Btu 


There has been no heat flow. The useful work equals the decrease of enthalpy, or 
1268.9 — 1123.4 = 145.5 Btu. Substituting in Eq. (16:13), 


Irreversibility = —145.5 — (—194.3) = 48.8 Btu 


Heat exchangers may, for our purpose, be divided into two subclasses, 
but for neither of these subclassifications is any significant amount of 
useful work performed; the main objective is the transfer of heat. Thus, 
in Eq. (16:13), i W 2 is zero, and we may ignore it in the calculation of 
irreversibility. The first kind of heat exchanger accomplishes its purpose 
by mixing two or more streams of fluid; ultimately, this flowing mixture 
may again divide into separate streams. If it is assumed that the only 
exchange of heat has been between the fluids of these streams, this can 
also be made an adiabatic process when we include in the system all of the 
streams. Thus dQ, as well as iW 2 , has a zero value in Eq. (16:13), and 
the calculation of the irreversibility of the process becomes a summing of 
the availabilities of all of the entering streams and of all the outgoing 
streams and a subtraction of the second total from the first. The relative 
mass rates of flow in the various streams must, of course, be taken into 
account in applying Eq. (16:13) and, for the heat exchanger, 

Irreversibility = £ M (* + S~g + l) ~ ^ M ( 6 + Sg + j) 

i - 0 

(16:14) 


in which ^ represents the summation of all incoming streams and 

i 

corresponding summation for those outgoing. 

The second class of heat exchangers keeps the streams of fluid separate 
as heat is passed between them through surfaces such as the walls of tubes. 
When Eq. (16:13) is applied to each stream separately, there will be a flow 
of heat, although no useful work is involved; the irreversibility, as calcu¬ 
lated by Eq. (16:13), will be zero. However, the total increase of avail¬ 
ability for those streams which are receiving heat will be less than the 
decrease of availability for the streams which are supplying this heat. 
Considering all streams to constitute a single system, the process is again 
an adiabatic as well as a no-work process, and Eq. (16:14) may be used to 
evaluate the irreversibility of the heat-exchange process.- 






THE EVALUATION OF IRREVERSIBILITY 


399 


Example 16:5(7. The steam exhausted from the turbine of Example 16:56 is 
condensed at constant pressure to a saturated liquid in a surface condenser. Cooling 
water enters this condenser at 70°F and leaves at 130°F. Its velocity may be assumed 
negligible. Per pound of wet steam entering the condenser, calculate (a) the change 
of availability and the irreversibility that accompanies the condensation of the steam, 
(6) the change of availability and the irreversibility that accompanies the heating of 
the cooling water, and (c) the irreversibility of the process as a whole. 

Solution: 

(a) From Example 16:5.6, the enthalpy h x of the steam as it enters the condenser is 
1123.4 Btu, and bi = 213.9 Btu. As a saturated liquid at the same pressure, h 2 = 
180.07 Btu and b 2 = 180.07 - (530) (0.3120) = 14.7 Btu. 

A (Availability) = b 2 — bi = 14.7 — 213.9 = —199.2 Btu 

In Eq. (16:13), i W 2 = 0 and / - ^ —- dQ may be evaluated on the basis of the 

constant temperature at which condensation took place, or 212 + 460 = 672°R. 
The total heat flow from the steam as it condensed was h 2 — h x = 180.07 — 1123.4 
= —943.3 Btu. Thus 


f* T ~ t — dQ = -943.3 (- 72 672 53 ° ) = -199.2 Btu 

and, from Eq. (16:13), 

Irreversibility = —199.2 — ( — 199.2) = 0 

(6) The heat absorbed by 1 lb of cooling water is equal to its change of enthalpy. 
Although this water actually carries an overpressure, the effect on its enthalpy is 
negligible and the values for saturated water will be used. Thus 

1 Q 2 = h/ 2 — hf l — 97.90 — 38.04 = 59.86 Btu per pound cooling water 

Weight of cooling water per pound of steam condensed = M = = 15.75 lb 

59.OO 

bi = 38.04 - (530)(0.0745) = -0.45 Btu; b 2 = 97.90 - (530)(0.1816) = 1.65 Btu 
A (Availability) = M (b 2 — bi) = 15.75[1.65 — ( — 0.45)] = 33.1 Btu 

r 2 j 7 y 7 

This process, considered alone, also shows no irreversibility, for if / -——- dQ is 

evaluated, it may be shown to have a value of 33.1 Btu and since 1 W 2 = 0, Eq. (16:13) 
gives 

Irreversibility = 33.1 — 33.1 = 0 

(c) For the process as a whole, there has been no transfer of energy to or from external 
systems either as heat or as work. The transfer of energy in this case is between parts 
of what is now the same system. The irreversibility of the process as a whole is equal 
to the decrease of availability of that system, or 

Irreversibility = —A (availability) = —( — 199.2 + 33.1) = 166.1 Btu 

As an alternate method, Eq. (16:14) may be used. 

Irreversibility = [(1)(213.9) + (15.75)(— 0.45)] - [(1)(14.7) + (15.75)(1.65)] 

= 166.1 Btu 

This irreversibility of the process as a whole reflects the effect of the finite difference in 
temperature between the steam and the cooling water. This temperature interval 
did not enter into the calculation when the two streams were considered individually. 







400 


BASIC ENGINEERING THERMODYNAMICS 


Apparatus which carries out a throttling process, as in the throttle- 
governing of engines or turbines or the expansion valve of the vapor- 
compression refrigerating plant, is sometimes used in engineering plants. 
These are no-heat no-work processes, and the irreversibility is equal to 
the decrease of availability. An illustration will be found in Example 
16:5A, part c. 

16:6. Efficiency versus Effectiveness of the Turbine Stage. The 

efficiency of a turbine stage has been defined in Art. 11:13 as the ratio of 
the actual enthalpy drop across the stage to the isentropic drop that 
would have been obtained if the expansion had been reversibly, as well 
as adiabatically, performed. In Fig. 16:1, a Mollier diagram, the expan- 



Fig. 16:1. Stage efficiency vs. stage effectiveness. 

sion in a high- and a low-pressure stage of the same real turbine has been 
represented by the lines ab and mn, respectively. The expansions in 
reversible stages from the same initial states a and m to the same final 
stage pressures pb and p n are shown as ac and mq. These have, for the 
purpose of convenient comparison in the analysis to follow, been made 
equal in length. Also, for the same reason the efficiencies of these two 
stages have been assumed to be the same. 

The efficiency of the high-pressure stage is ( h a — hb ) / ( h a — h c ), and the 
work performed in that stage is h a — hb, per pound of fluid flow; this 
assumes that there is no significant difference in velocity or elevation of 
the fluid between stage entrance and exit. At first thought, the irreversi¬ 
bility of this stage would seem to be equal to the unrealized drop of 
enthalpy, h b — h c . But the fluid leaves the real stage at state b , the 
isentropic stage at state c. The isentropic, or ideal, expansion to con¬ 
denser pressure p n from state b will produce work in the amount hb — hb>, 
while an expansion from state c to the same pressure will develop h c — h c >, 
a smaller amount of work because of the divergence of constant-pressure 








THE EVALUATION OF IRREVERSIBILITY 


401 


lines pb and p n . In other words, the decrease of availability has not been 
as large for the real stage as for the isentropic stage, and although the 
work in the real stage is less than the decrease of availability in that stage 
while, in the isentropic stage, these two quantities are equal, the picture 
is not quite as dark as it is painted by the value which we call the stage 
efficiency; some of the apparently lost work hb — h c is recoverable in later 
expansion. 

The effectiveness of a turbine stage gives a more accurate picture of the 
effect of the irreversibilities encountered in a single turbine stage on the 
performance of the turbine as a whole. Effectiveness is expressed as the 
ratio the work performed in the real stage bears to the decrease of availa¬ 
bility that has taken place in that stage. The work of the high-pressure 
stage of Fig. 16:1 is h a — hb’, making the same assumption with regard to 
equality of fluid velocity and elevation at entrance to and exit from the 
stage that has been the basis of the accepted expression for stage efficiency, 
the decrease of availability in this stage is h a — hb, and the effectiveness of 
the stage is (h a — h b )/(h a — h b ). This may be compared with (h a — hb)/ 
(h a — h c ), the stage efficiency, as follows: 

h a — hb — (h a — ToS a ) — (hb — ToSb) = h a — hb To(sb — s a ) 

In the real stage, Sb is always greater than s a , and T 0 cannot be negative; 
thus T 0 (sb — s a ) is always a positive quantity. If this quantity is smaller 
than the positive difference of enthalpy, h b — h c , then (h a — h b ) < (h a — h c ) 
and the stage effectiveness will exceed the stage efficiency. The differ¬ 
ence of enthalpy h b — h c is equal to the heat flow required to change the 
state of unit weight of the fluid from c to h at constant pressure p b and 

therefore to T ds, where T is the (possibly variable) temperature along 

the constant-pressure path cb. If the temperature between the points c 
and h is everywhere above T 0 , as it always is in any stage of a real turbine, 

then f b T ds > T Q (s b - s c ). But s c = s a and T 0 (s b - s a ) may be sub¬ 
stituted for To(s b — s c ) in this inequality; also, hb — h c may substitute for 
T ds, and the inequality becomes (h b — h c )> To(s b — s a ). The effec¬ 
tiveness of any real stage will therefore exceed its efficiency. The con¬ 
ventional expressions for stage efficiency and stage effectiveness are 

( 16 : 15 ) 

( 16 : 16 ) 

where t? s and e s are, respectively, the stage efficiency and effectiveness, 
-Ah is the drop of enthalpy that takes place in the real stage, -A h s is the 


Vs = 


e s = 


— Ah 
— A h 8 
— Ah 
^A h 




402 


BASIC ENGINEERING THERMODYNAMICS 


isentropic drop of enthalpy between the entrance state and the exit pres¬ 
sure, and — A b is the decrease of availability between states at entrance 
to and at exit from the real stage. It is also possible to state that, for 
any real stage, e s > rj s . 

Having developed the concept of stage effectiveness from a study of the 
high-pressure stage of Fig. 16:1, let us apply it to the low-pressure stage 
of that figure. It will be assumed that this stage exhausts to the con¬ 
denser, and the initial assumption will also be that the temperature main¬ 
tained in constant-pressure condensation in the condenser is the tempera- 

ture T o of the atmosphere. In that case it will be noted that J T ds = 

T 0 (s n — s m ), that b m — b n = h m — hq, and that the efficiency and the 
effectiveness of the stage are therefore equal. Since the efficiencies of the 
high- and low-pressure stages are the same by assumption and the effec¬ 
tiveness exceeded the efficiency in the high-pressure stage, this means 
that the effectiveness of the low-pressure stage is less than that of the 
high-pressure stage. Although the condenser temperature can never be 
as low as T 0 in practice and thus the effectiveness will always be greater 
than the efficiency, it is evident that as the temperature at exit from a 
stage declines toward the temperature of the atmosphere the difference 
between the efficiency and the effectiveness will decrease; this decrease 
will result from an approach of —A b of Eq. (16:16) toward the —A h s of 
Eq. (16:15). 

The irreversibility of a turbine stage is the difference between the drop 
of availability in the stage and the work of the stage, or —A b — ( — Ah) = 
Ah — Ab. It has been shown that —A b becomes a larger fraction of 
— A h s for a stage of given efficiency when the exhaust temperature from 
the stage is low. Therefore the irreversibility of a low-pressure stage will 
be greater than that of a high pressure stage of equal efficiency and having 
the same isentropic enthalpy drop. We may therefore draw the impor¬ 
tant conclusion that low efficiency in the high-pressure stages of a turbine 
has a less serious effect on the efficiency of the turbine as a whole than if 
the low-pressure stages operate at low efficiency. This conclusion is 
reflected in the design of real turbines, in which more attention is usually 
directed toward the reduction of losses in the final stages of expansion 
than in the early stages. 

When the condition line (see Art. 11:14 and Figs. 11:17 and 16:1) is 
plotted for a turbine that has stages of equal efficiency, a curve will result 
on the Mollier diagram as shown in Fig. 16:1; if the effectiveness of each 
stage is the same, the condition line will be straight. In Art. 11:14 it was 
shown that the efficiency of a multistaged turbine which has equal stage 
efficiencies will differ from the efficiency of its stages; on the other hand, 
if the effectiveness of all stages is the same, the over-all effectiveness of 
the turbine will be identical with that of each of its stages. 


THE EVALUATION OF IRREVERSIBILITY 


403 


Example 16:6. Steam enters the first stage of a turbine at 600 psia, 700°F, and 
leaves the stage at a pressure of 240 psia. The stage efficiency is 70 per cent. The 
last stage of this turbine receives steam at 5.5 psia, x = 0.944, and exhausts to the 
condenser pressure of 1 psia. The efficiency of this stage is also 70 per cent. Atmos¬ 
pheric pressure and temperature are 14.7 psia and 70°F. Velocities at entrance to 
and exit from each stage will be neglected. Calculate the enthalpy drop, the decrease 
of availability, the effectiveness, and the irreversibility for (a) the high-pressure stage 
and (6) the low-pressure stage. 

Solution. For convenience, the notation of Fig. 16:1 will be used. From the 
Mollier diagram the following information may be obtained: 

h a — 13ol; s a — 1.58/ = s c ; h c — 1251 
h m = 1077; s m = 1.749 = s 9 ; h q = 977 

(a) -Ah = 0.70 (h a - h c ) = 0.70(1351 - 1251) = 70 Btu/lb 

h b = 1351 - 70 = 1281 


From the Mollier chart, at 240 psia this corresponds to s& = 1.617. 


ha = 1351 - (530)(1.587) = 510; b b = 1281 - (530)(1.617) = 424 

Decrease of availability = -A(availability) = -Ab = 510 - 424 = 86 Btu/lb 


€s 


— Ah _ 70 

— Ab ~ 86 


0.815 


Irreversibility = Ah - Ab = - 70 -f- 86 = 16 Btu per lb 


(b) -Ah = 0.70 {h m - h q ) = 0.70(1077 - 977) = 70 Btu 


h n = 1077 - 70 = 1007 Btu 
corresponding, at 1 psia, to s n = 1.802 


b m = 1077 - (530)(1.749) = 150; b n = 1007 - (530)(1.802) = 52 
— Ab = 150 - 52 = 98 Btu/lb 
0.715 

Irreversibility = — 70 -f 98 = 28 Btu/lb 


It will be observed that the data for the two stages have been so selected that they 
have equal enthalpy drops as well as equal efficiencies; this permits a direct com¬ 
parison of their irreversibilities. 


Problems 

In the problems to follow, the temperature of the atmosphere will be assumed to be 
60°F, its pressure 14.7 psia, except when specifically stated otherwise. 

1. Find the availability of 1 lb of H 2 0 at a pressure of 100 psia in the following 
conditions: (a) t = 600°F; (b) saturated vapor; (c) saturated liquid; ( d ) t = 60°F; 
(e) t = 32°F. 

2. Find the availability of 1 lb of H 2 0 at a temperature of 500°F in the following 
states: (a) p = 50 psia; (6) p = 14.7 psia; (c) saturated vapor; ( d ) saturated liquid; 
(e ) p = 500 psia. 

3. Find the availability of 1 lb of air at the following states. Assume the internal 
energy and the entropy of the air to be referred to (to be zero at) a state in which it 
is in temperature and pressure equilibrium with the atmosphere, (a) p = 100 psia, 
t = 60°F; (6) p = 100 psia, t = 0°F; (c) p = 100 psia, t = 200°F; (d) p = 100 psia, 
v = 2 ft 1 2 3 /lb; (e) p = 14.7 psia, t = 60°F; (/) p — 14.7 psia, t = 0°F; ( g ) p = 14.7 
psia, t = 200°F; (h) p = 2 psia, t = 60°F; (i) p = 2 psia, t = 0°F; ( j) p = 2 psia, 



404 


BASIC ENGINEERING THERMODYNAMICS 


t = 200°F. Would your answers have differed if the internal energy and the entropy 
had been referred to a different state? 

4. Plot a 13s chart for steam. Show (a) the saturated-liquid and saturated-vapor 
lines and the critical point; ( b ) lines of constant pressure of 4000 psia, 2000 psia, 
14.7 psia, and 0.25 in. Hg; (c) lines of constant temperature of 800, 600, 60, and 32 F; 
(d) the triple-point region. 

5. Show the appearance of a closed-system (nonflow) Rankine cycle on the chart 
of Prob. 4. Assume that p u = 200 psia, pi = 14.7 psia and (a) that the steam is 
saturated at the end of the constant-pressure expansion and (b) that its temperature 
is 500°F at this point in the cycle. 

6 . Work Example 16:3 as based on an atmospheric temperature of 60°F. Com¬ 
paring with the results of the example, do you find any difference in the changes of 
availability that take place during the listed processes? Any changes in the calcu¬ 
lated irreversibilities? Explain. 

7. Work Example 16:3, changing the original state to a saturated vapor at 100 psia 
and the temperature of the atmosphere to 60°F. 

8 . The system consists of 1 lb of saturated steam at 100 psia. Calculate the change 
of availability and the irreversibility that accompany each of the following processes: 
(a) While confined in a rigid-walled container, the temperature of the system is 
raised to 500°F by the slow addition of heat. (6) The same change of state as in 

(a) is effected adiabatically, as the result of a paddle-wheel process, (c) The system 
expands adiabatically into an adjoining space which is initially empty. The final 
volume is twice the initial volume. ( d ) The system is originally in turbulent motion, 
storing kinetic mechanical energy in the amount of 2 Btu. As the result of friction, 
it comes to rest while still confined within the same rigid-walled container. 

9. The system consists of 1 lb of air at 100 psia and 300°F. Calculate the change 

of availability and the irreversibility that accompany each of the following processes. 
Assume the internal energy and the entropy of the air to be zero at a state in which 
it is in temperature and pressure equilibrium with the atmosphere, (a) The system 
expands slowly and at constant pressure behind a piston as heat is supplied to it. 
The final temperature is 400°F. (6) The same state path as in (a) is followed but 

as the result of an adiabatic process, (c) The system expands adiabatically and 
reversibly to a final pressure of 14.7 psia. ( d) The system expands adiabatically to 
a final pressure of 14.7 psia and final temperature of 60°F. (e) The system expands 

slowly and reversibly behind a piston at constant temperature to a pressure of 14.7 
psia. 

10. The system consists of one pound of air at 70 psia and 200°F and a second 
pound of air at 15 psia and 60°F. They are contained in separate compartments of 
a tank having rigid nonconducting walls. The partition that separates them is 
punctured, allowing the two bodies of air to mix and reach a common pressure and 
temperature. Calculate the change of availability of the system and the irreversi¬ 
bility of the process. 

11 . A pound of nitrogen and a pound of oxygen, both at 14.7 psia and 60°F, are 
confined in separate compartments of a container with rigid nonconducting walls. 
The partition between the two is punctured, allowing the two gases to mix. Calcu¬ 
late the irreversibility of the process. 

12. In Prob. 10, assume the partition that separates the two bodies of air to be 
rigid but to allow the passage of heat. The partition is not punctured, but the two 
bodies of air reach a common temperature, (a) For the system as described in 
Prob. 10, calculate the change of availability and the irreversibility of the process. 

(b) For a system composed of the pound of air that was originally at 70 psia, calcu¬ 
late the same quantities, (c) For the system composed of the pound of air originally 


THE EVALUATION OF IRREVERSIBILITY 


405 


at 15 psia, calculate the same quantities. ( d ) Compare your answers to parts b and c 
with your answer to part a, and explain any apparent inconsistency. 

13. Plot a bs chart for steam. Show (a) the saturated-liquid and saturated-vapor 
lines and the critical point, (6) lines of constant pressure of 4000, 2000, and 14.7 psia 
and 0.25 in. Hg abs, (c) lines of constant temperature of 800, 600, 60, and 32°F, and 
(d) the triple-point region. 

14. Show the appearance of a steady-flow Rankine cycle based on 1 lb of H 2 0 on 
the chart of Prob. 13. Assume that p u = 200 psia, p\ = 14.7 psia, and (a) that the 
steam is saturated as it enters the prime mover and (6) that its temperature is 500°F 
at this point in the cycle. 

15. Work Example 16:5A, changing the temperature of the atmosphere to 60°F. 
Compare with results of the example as solved in the text, and state your conclusions. 

16. Work Example 16:5A, changing the original state of the H 2 0 to a saturated 
vapor at 100 psia and the temperature of the atmosphere to 60°F. 

17. A steady stream of air at 2 psia and 60°F enters a process at negligible velocity. 
What is its availability per pound? During the process, it expands adiabatically 
and reversibly to a pressure of 1 psia. At what velocity does it leave the process? 
Calculate the change of availability per pound. Show that the irreversibility of the 
process is zero. 

18. A steady stream of H 2 0 enters a process at 100 psia, 400°F, and at a velocity 
of 100 fps. Calculate the change of availability and the irreversibility, per pound of 
H 2 0, for each of the following processes: (a) The stream is heated slowly at constant 
pressure until its temperature is 500°F. The passage has a constant cross-sectional 
area and is horizontal. (6) As it passes through a horizontal pipe of constant diam¬ 
eter, it is heated to 500°F. Because of friction, the final pressure is 90 psia. (c) It 
expands adiabatically and reversibly through a nozzle to a pressure of 14.7 psia. 
( d ) It expands adiabatically through a nozzle which it leaves at 14.7 psia and a 
quality of 0.95. ( e ) It is throttled slowly and adiabatically to a pressure of 14.7 psia. 
The velocity at exit differs negligibly from the initial velocity. (/) It expands 
adiabatically to 14.7 psia as it passes through a turbine with an efficiency of 55 per 
cent. 

19. The stream of air described in Prob. 17 enters a compressor which raises its 
pressure to 14.7 psia. It leaves the compressor at a velocity of 100 fps. Calculate 
the change of availability per pound in the compressor and the irreversibility of the 
compressor process, based on the following assumptions: (a) The compression is 
adiabatic and reversible. (6) The compression is isothermal and reversible, (c) The 
final temperature of the air is 600°F, and the compression is adiabatic. ( d ) The 
compression is adiabatic, and the compressor has an adiabatic efficiency of 80 per cent. 

20. In Prob. 19, let the air enter the compressor at 14.7 psia and 60°F at negligible 
velocity. It leaves at 100 psia, also at negligible velocity. Calculate the change of 
availability and the irreversibility of the process, per pound of air, for each of the 
processes described in parts a to d of Prob. 19. 

21. In Example 16:55, assume the temperature of the atmosphere to be 60°F, and 
calculate the same quantities. Compare with the answers of the example, and state 
your conclusions. 

22. In Example 16:55, let p 0 = 14.7 psia and t 0 = 70°F as in the example, but 
assume that the steam is expanded to a final pressure of 1 psia. Compare your 
answers with those of the example, and state your conclusions. 

23. A steady stream of exhaust steam enters a surface condenser at 1 psia and a 

quality of 0.95 at negligible velocity. It leaves as a saturated liquid at the same 
pressure and at negligible velocity as the result of an energy transfer with a steady 
stream of cooling water which enters at 60°F and leaves at 80°F. (a) Calculate the 


406 


BASIC ENGINEERING THERMODYNAMICS 


weight of cooling water circulated per pound of steam condensed. ( b ) Considering 

1 lb of condensing steam as the system, what is the change of availability and the 
irreversibility that accompany the process? (c) What are the similar quantities 
when the system is the weight of cooling water that is required to condense 1 lb of 
steam? ( d ) For a system composed of 1 lb of condensing steam plus the weight of 
cooling water that must be circulated to bring about its condensation, what are the 
change of availability and the irreversibility of the process? ( e ) Compare the 
answers to (6) and (c) with the answer to part d, and explain any apparent discrepancy. 

24. In Example 14:4, 0.119 lb of steam at 225 psia and a specific enthalpy of 
1123 Btu enters the high-pressure regenerative feedwater heater and is condensed 
to a saturated liquid at the same pressure while 0.881 lb of feedwater at 600 psia and 
294.8°F (h = 264.3) enters from the second heater and is raised at constant pressure 
to a temperature of 391.8°F (h = 366.1). What is the irreversibility of the process 
that takes place in this heater? Assume velocities to be negligible. 

25. In Example 14:4, calculate the irreversibility of the processes that take place 
in (a) the intermediate heater and (6) the low-pressure heater, per pound of steam 
supplied the turbine by the boiler. 

26. In a steam power plant that operates condensing at an exhaust pressure of 

2 psia, a part of the steam is used to supply steam-driven auxiliaries such as the feed- 
water pump, blowers, etc. The exhaust from these auxiliaries is available at atmos¬ 
pheric pressure and can be used to preheat the feedwater in an open feedwater heater, 
in which it mixes directly with the feedwater. The exhaust steam is saturated as it 
enters the heater; the feedwater enters at 100°F, and the mixture of feedwater and 
condensate leaves at 210°F. Per pound of feedwatei entering the boiler, i.e., per 
pound of the mixture, calculate the irreversibility of the process. 

27. A throttling governor throttles the saturated steam leaving the boiler at a 
pressure of 150 psia to a pressure of 100 psia before it enters the prime mover. Per 
pound of steam, calculate the irreversibility of the process carried out by the governor. 

28. In an economizer, the flue gases ( c v = 0.24) are used to preheat the feedwater 
entering the boiler of a steam power plant and, in doing so, decrease in temperature 
from 630 to 400°F. The feedwater is heated from 210 to 290°F. Find, per pound of 
feedwater entering the boiler, (a) the weight of flue gas cooled and ( b ) the change of 
availability and the irreversibility of the process carried out in the economizer. 

29. In an air preheater, the ratio of the mass rate of flow of flue gas (c p = 0.24) to 
that of the air for combustion is 1.05. Fourteen pounds of air is supplied for the 
combustion of each pound of coal fired, and 10 lb of steam is generated per pound of 
coal. The air for combustion is heated from 60 to 210°F. The flue gas enters the 
preheater at 580°F. Per pound of steam generated, what are the change of availa¬ 
bility and the irreversibility of the process that takes place in the air preheater? 

30. In Example 11:14, calculate the effectiveness of each of the four stages. What 
data must be introduced other than those included in the statement of the example? 

31. Use the data of Example 11:14, but assume equal stage effectivenesses of 
70 per cent. ( a ) Calculate the efficiency of each stage and the efficiency of the 
turbine. (6) Calculate the effectiveness of the complete turbine, (c) Show that 
the condition curve is a straight line. 

32. Use the data of Example 11:14, but assume stage effectiveness to be as follows: 
first stage, 80 per cent; second stage, 75 per cent; third stage, 70 per cent; fourth 
stage, 65 per cent, (a) What is the condition of the steam as it leaves the turbine 
(neglect velocity)? (6) Calculate the efficiency of each stage and of the turbine, 
(c) Calculate the effectiveness of the turbine, and show that it equals an average of 
the effectivenesses of the individual stages when that average is weighted according 
to the respective decreases in 6 in the stages. 


THE EVALUATION OF IRREVERSIBILITY 


407 


33. (a) Prove the statement made in the text to the effect that if the effectiveness 
of each stage of a turbine is the same, the over-all effectiveness of the turbine will be 
identical with that of each of its stages. ( b ) Prove that the condition line is a straight 
line for a turbine having stages of equal effectiveness, (c) Prove that the effective¬ 
ness of a turbine is an average of the effectivenesses of its stages when that average 
is weighted according to the decrease in availability in each stage. 

34. (a) A turbine stage receives steam at 400 psia, 600°F, and exhausts at 200 psia. 
Its effectiveness is 72 per cent. Calculate the enthalpy drop, the decrease of availa¬ 
bility, the efficiency, and the irreversibility per pound of steam in this stage. Neglect 
velocities. 

(6) The last stage of this turbine exhausts wet steam at a pressure of 1 in. Hg abs 
and a quality of 0.85. The isentropic enthalpy drop for this stage is the same as for 
the stage described in part a, and its effectiveness is also 72 per cent. What is the 
condition of the steam as it enters the last stage? What is the actual enthalpy drop 
in this stage? What are the decrease of availability and the irreversibility per pound 
of steam in this stage? What is the stage efficiency? Neglect velocities. 

• 

Symbols 

b h — T 0 s, a part of the availability function of unit mass in steady flow 
c p specific heat at constant pressure 
e stored energy of a system of unit mass 
E stored energy of a system 
g acceleration of gravity 
h specific enthalpy 
J proportionality factor 
M mass rate of flow 
p pressure, psi 

P pressure, psf; pressure in general 
P 0 pressure of the atmosphere 
Q heat flow; rate of heat flow 
$ specific entropy 
S entropy of a system 
T absolute temperature 
To absolute temperature of the atmosphere 
u specific internal energy 
v specific volume 

V volume of a system 

V velocity 

W work; rate of work delivery 
x quality of a vapor 
z elevation 

Greek Letters 

(3 availability function of a simple system of unit mass 
e s effectiveness of a turbine stage 
Vs efficiency of a turbine stage 

Subscripts 

0 at the dead state 
s stage 
u useful 


CHAPTER 17 


THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 

17:1. Introduction. In Chap. 11, in discussing the flow of fluids 
through nozzles, the basic assumption was made that the flow was adia¬ 
batic and reversible. Although this assumption was never quite justified, 
the deviations from reversibility in flow were relatively small and could be 
conveniently expressed in terms of so-called “velocity ” and “discharge’’ 
coefficients having values usually very close to 1. 

If we apply the assumption that the flow is reversible as well as adia¬ 
batic to the flow of a fluid through a channel of constant cross-sectional 
area, such as a pipe, the following flow conditions would follow: 

1. The state of the fluid, including its velocity, would be identical at 
all points along the passage. 

2. Since there is no differential of pressure, no power would be required 
to transport the fluid through a channel of any length. 

3. The mass rate of fluid flow would depend only on its specific volume 
and its velocity, both of which would be the same at all sections of the 
channel, and therefore would be independent of the length of the channel. 

The conditions for the adiabatic flow of real fluids through long pipes 
and ducts are readily recognized as bearing no resemblance to those out¬ 
lined above. The pressure will decrease in the direction of flow, often 
with an increase in specific volume and velocity as a consequence, power 
will be required for the transport of the real fluid, and the mass rate of 
flow will depend on the differential of pressure that is maintained between 
entrance and exit and upon the length of the passage. The cause of the 
irreversibility is found in the property of the real fluid that is called 
viscosity. Because of its viscosity, the fluid tends to be slowed by its 
contact with the walls of the channel. But the same mass rate of flow 
must be maintained past each channel section, just as in flow through a 
nozzle, and a decrease of pressure in the direction of flow becomes neces¬ 
sary to maintain the required velocity. In the case of compressible 
fluids, this pressure differential is still further increased by the increase in 
specific volume that accompanies reduced pressure, making it necessary 
that the velocity be not just maintained at the same value for successive 
sections but increased in proportion to the increase in specific volume. 

The force applied by the walls of the channel on the column of fluid, 
due to the latter’s viscosity, can act only to decrease the velocity of the 

408 


THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 409 


fluid relative to these walls; the action of viscosity is thus to create an 
irreversibility. It is only because of the relatively short length of a nozzle 
that the flow through it may be assumed to approach reversibility. 

There have been many occasions in earlier chapters when reference, 
direct or implied, was made to the viscosity of the real fluid. For exam¬ 
ple, the paddle-wheel process first discussed in Art. 1:7 was a possible 
means of introducing energy into the system only because that system 
was composed of a real fluid with viscosity as one of its properties. Pre¬ 
vious references have, however, been of a general character, and no 
attempt has been made to measure the forces that are set up as the result 
of viscosity. Since viscosity is an important property in connection with 
the flow of real fluids through long pipes and ducts, that discussion can no 
longer be delayed. 



Fig. 17:1. Viscosity and velocity gradient. 


17:2. Viscosity and Viscous Drag. Viscosity is the resistance that 
the fluid offers to shearing motion. Consider, as in Fig. 17:1, two parallel 
flat plates separated by a slice of fluid having the thickness y. The lower 
plate is anchored and remains stationary; the upper plate, of area A, is 
kept in slow motion at a velocity V through the agency of a force F which 
acts in the plane of the plate; the magnitude of this force is a measure of 
the viscous drag. 

Experiment has shown that a thin layer of the fluid will remain attached 
to each plate, the layer attached to the lower plate thus being stationary 
and that attached to the upper having the velocity V. Intervening 
laminae of fluid will move with intermediate velocities which will be 
greater as their distance from the lower plate increases. If the velocity 
is slow, the particles of fluid in these laminae will follow nonintersecting 
(streamline) paths, and the motion of the fluid may be described as 
laminar or streamline, or since the force exerted by a lamination in slowing 
or accelerating more or less rapidly traveling adjacent laminae is due 
solely to the effects of viscosity, the term viscous flow is also used. When 
the velocity exceeds a certain critical value which will be shown later to 
depend on the distance between the plates and the density and viscosity 












410 


BASIC ENGINEERING THERMODYNAMICS 


of the fluid, particles will begin to cross paths and to collide with each 
other; the flow is now beginning to be turbulent. In the present article 
we shall confine ourselves to a discussion of laminar flow; a discussion of 
the effects produced in turbulent flow will be taken up later. 

In Fig. 17:1 is shown a velocity profile that plots the velocity of each 
fluid lamination with respect to its position between the two plates; in 
general, unless the distance between the plates is very small, this will be a 
curve as shown. The slope dV/dy of this curve is therefore not constant. 
Experiment has shown that the force F will be in proportion to the slope 
of this curve in the laminations immediately adjacent to the plate to 
which the force is applied, i.e., to the ratio (dV/dy) 0 shown in the figure; 
this is called the velocity gradient. The force F will also depend on the 
area A of the upper plate. This area is the “ wetted ” surface; if the plate 
is submerged, the area of both sides must be considered, each, of course, 
being used in conjunction with its respective value of ( dV/dy) 0 . Another 
factor in the size of F is the viscosity of the fluid, which will be designated 
as It is the arbitrary practice to assign a value to u which is in direct 
proportion to F under equivalent conditions as to area and velocity 
gradient. These are the only factors that control the size of F in laminar 
flow, and we may write an expression for the resistance R (force per unit 
area of wetted surface) in the form 



in which u is a dimensional constant that expresses the relative viscosity' 
and is called the coefficient of viscosity of the fluid. Since the coefficient of 
viscosity is not dimensionless, as an examination of Eq. (17:1) will show, 
the value it has in expressing a given viscosity will vary with the units in 
which F, A, V , and y are measured. When the unit of force is the dyne, of 
distance is the centimeter, and of time is the second, the corresponding 
unit of viscosity is called the poise. It has become usual practice to 
express the viscosity in tables, equations, and charts in terms of the centi- 
poise, or T ^o poise. 

In general, the viscosity of a fluid is practically independent of its pres¬ 
sure 1 and is primarily a function of its temperature. If the fluid is a 
liquid, its viscosity will decrease with increased temperature; if a gas, 
increased viscosity will accompany increased temperature. Table 17:1 
gives approximate equations for the viscosity, in centipoises, of a few 
gases and liquids. These equations are valid only in the approximate 

1 This is true except for a vapor at a temperature that is close to its boiling point. 
For a few vapors (see, for example, Keenan and Keyes, “Thermodynamic Properties 
of Steam,” Table 6, p. 76, John Wiley & Sons, Inc., New York, 1936) the effect of 
pressure on viscosity has been investigated. 




THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 411 


range 0 to 200 F. They show a straight-line relation between viscosity 
and temperature; the actual relation is not nearly so simple, but these 
equations are presented for the convenience of the reader in the solution 
of problems to follow and to illustrate the typical manner in which viscos¬ 
ity increases with increase of gas temperature, or decrease of liquid 
temperature. 


Table 17:1. Viscosity of Gases and Liquids 
t in °F, viscosity in centipoises 


Substance 

Equation 

Substance 

Equation 

Gases 

Air. 

M = 0.0165 + 2.5 X 10~5< 

Liquids 

Gasoline, sp. gr = 0.7. 

M = 0.57 - 0.0021* 

Oxygen. 

M = 0.018 + 2.5 X 10-5f 

Lubricating oils 


Nitrogen. 

M = 0.0155 + 2.5 X 10-5* 
n = 0.0078 + 1.25 X 10- 6 < 

M = 0.0133 + 2.5 X 10-5* 
ix = 0.0083 + 2 X 10-5* 

SAE 10 

n — 160 — * 
ix = 649 - 4.15* 
ix = 1.80 - 0.003* 

Hydrogen. 

SAE 30 

Carbon dioxide. 

Mercury. 

Steam. 

Water 

Methane. 

ix = 0.0095 + 1.5 X 10-5* 

32-100°F. 

ix = 2.317 - 0.0164* 
n = 1.027 - 0.0035* 


100-212°F. 


Viscosity is a property of the fluid. Thus, although the value given \x 
is based on the resistance to laminar flow, this value will also describe the 
viscosity of the same fluid at the same state, even when the flow is 
turbulent. 

The equation presented in Table 17:1 for the calculation of the viscosity 
of steam applies only at low pressure. A much more complete and accu¬ 
rate picture in respect to the viscosity of steam and of water is found in 
Table 6 of the Keenan and Keyes steam tables. In Table 6, the unit of 
viscosity is based on the pound of force, the second, and the foot (lb-sec/ 
ft 2 ). This unit of viscosity corresponds to 47,800 centipoises. 

17:3. Dimensional Analysis. Even when the value of /z, the coefficient 
of viscosity, is known, Eq. (17:1) offers little direct assistance to the 
engineer in the calculation of the effect of fluid viscosity on the flow 
through a long pipe or duct. Before the magnitude of the force that acts 
to slow the column of fluid may be computed, he must be able to assign a 
value to the velocity gradient if that equation is used, and that velocity 
gradient is affected by a number of factors such as the density of the fluid, 
the average velocity of flow in the channel, and the size (diameter) of the 
pipe as well as the viscosity of the fluid. Moreover, as has been stated, 
Eq. (17:1) is valid only for low velocities. In order to express the way in 
which the various factors interact and to determine their combined effect 
on the resistance to flow, the engineer uses a tool which he calls dimen¬ 
sional analysis. 

The basic dimensions with which the engineer is concerned include mass 























412 


BASIC ENGINEERING THERMODYNAMICS 


(. M ), length, or distance (L), and time ( 6 ). x There are many units of 
mass with which the engineer deals; the pound, the slug, and the gram are 
examples. We may write 


1 slug = 32.2 lb mass 

This equation is based on the definition of the pound mass as that mass 
which weighs one pound if placed in the earth’s gravitational field. On 
the other hand, it would be dimensionally incorrect to write the preceding 
relation as “1 slug = 32.2 lb weight ,” for it is only in a gravitational field 
of an intensity equal to that of the earth’s that 1 slug would weigh 32.21b. 

A large number of quantities, such as force, weight, mass, velocity, 
elevation, acceleration, moment, work, heat, energy, etc., enter into 
engineering calculations, and there are various units in which the amounts 
of each may be measured. But each of these quantities may be expressed 
in terms of the basic dimensions M, L, and 6 without regard to the units 
in terms of which its measurement is to be made. Thus, to list only the 
few on which our interest is, for the moment, centered: 


Mass = [M] 
Length = [L] 
Time = [0] 
Area [A] = [L 2 ] 
Volume [V] = [L 3 ] 

Mass density [p] = 
Velocity [V] = 


V 

L 

e 2 


Acceleration 


dV 

d6 


M 

U 


Force (mass X acceleration) [F] 


ML 

e 2 


In each of the foregoing expressions, the brackets are used to identify the 
equality as being one of dimensions rather than, necessarily, of units of 
measurement. For example, the dimensional expression for velocity 
above may be read as “ The dimension of velocity is equal to the dimension 
of length divided by the dimension of time.” We shall add one more to 
this short list as being necessary to our present purpose. To express the 
viscosity dimensionally, we note from Eq. (17:1) that 

1 Force (F), length (L), and time (0) may also be used. In that case, the dimensions 
of mass would be expressed in terms of the dimensions of force, length, and time. It is 
sometimes convenient to add a temperature dimension ( T). 













THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 413 


or 


M = 


F dy 
A dV 



r ml \ 

[L] 

r i ' 




[Ml 

_ e 2 

L 2 _ 


L 

— 

17e_ 


To be correct, an engineering equation must be dimensionally con¬ 
sistent, i.e., the dimensions on the two sides of the equality sign must 
agree. This principle is often used to check the correctness of an equa¬ 
tion ; conversely, it may be used to set up a relation that is dimensionally 
correct. In this case it is our desire to formulate an equation which will 
express the resistance offered to the flow of a fluid through a pipe or con¬ 
duit. This expression, to be accurate and usable, should reflect the 
effects of the various factors which our analysis of the situation has indi¬ 
cated will affect the amount of this drag. These factors, as they have 
been listed above, include the mass density of the fluid, its viscosity, its 
velocity of flow in the channel, and the size of the channel. Thus we may 
write 

R = 2 = /(p,m,f,d) (i) 


in which R is the resistance to flow per unit area of contact with the flow¬ 
ing fluid (per unit area of wall surface), p is the mass density of the fluid, 
p is its viscosity and V its velocity, and D is a dimension of the channel, 
conventionally the diameter if the channel is round in cross section. 

The degree to which each of the quantities p, p, V, and D individually 
affects the resistance to flow is unknown but may be expressed by assign¬ 
ing to each an unknown exponent. Thus (1) may be written in the form 

R = Cp a n b V c D d (2) 


in which C is a dimensionless constant of proportionality and a, b, c, and d 
are the exponents expressing the unknown parts which the quantities to 
which they are respectively attached play in accounting for the resistance 
to flow. This equation may be written dimensionally in the form 


[LY (3) 

Noting that the dimensions must be consistent, we may write 


[R] = 


F 


ML 


1 


M 


M 

a 

M 

b 

L 

A 


d 2 


L 2 


Ld 2 


L 3 


Ld 


d 


Based on the mass dimension: 
Based on the length dimension: 
Based on the time dimension: 


a + b = 1 

—3 a — b c d — —1 (4) 

— b — c = —2 


There are thus three equations and four unknowns ( a,b,c,d ) so that, in 
solving, values of the unknowns must be expressed in terms of one of their 






























414 


BASIC ENGINEERING THERMODYNAMICS 


number. Observing that b is the only term that appears in every one of 
the group (4) equations, it is selected for this use, and we obtain 

a = 1 — b 

c = 2 - b (5) 

d = 3(1 - b) + b - (2 - b) - 1 = -b 

These values may be substituted in (2), giving 

R = Cp'^ b n b V‘ t ~ h D~ h = CpV 2 (6) 

A test will show that the product pVD/i u is dimensionless; it is known as 
the Reynolds number. Since it is dimensionless it may be combined with 
the dimensionless constant C to give a coefficient of friction f, 

, - C (f?y (17:2) 

indicating that this coefficient of friction, though dimensionless, is not 
constant but is some unknown function of the Reynolds number. Return¬ 
ing to (6), we may write 

R = fpF 2 (17:3) 

Dimensional analysis carries no further than to this point. To deter¬ 
mine what relation exists between f and Re, the Reynolds number, we 
must resort to experiment. Moreover, the results of dimensional analysis 
can be no more accurate than has been our analysis of the factors that 
affect the desired quantity. For instance, in the above example of 
dimensional analysis the factors chosen were four in number, while the 
number of dimensions was three; this produced a single dimensionless 
number on which the coefficient of friction will depend. But, in Chap. 
11, it is indicated that the velocity of sound is also a factor in the high¬ 
speed flow of gases and vapors. If the velocity of sound, V s , had been 
included, the number of unknowns in the group (4) equations would have 
been five and two dimensionless quantities, instead of one, would have 
resulted. It may be shown that one of the two would have been the 
Reynolds number, the other the ratio of fluid velocity to the velocity of 
sound in the fluid, or V/V s . This is the Mach number , a ratio of consider¬ 
able importance in studying the flow of gases or vapors at velocities close 
to the velocity of sound. At low Mach number the influence of Reynolds 
number is predominant; it is the Reynolds number that is of principal 
importance in the study of fluid flow through pipes and ducts. 

Example 17:3. Add the velocity of sound in the fluid to the factors assumed in 
Art. 17:3 to affect the resistance to flow. Show that the coefficient of friction will in 
that case depend on the Mach number as well as the Reynolds number. 




THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 415 


Solution: 

R = f(p,HjV,D,V t ) 

R = Cp a n b V c D d Vs e 

[SHmsrraM 5 .]' 

Based on the mass dimension: a + b = 1 

Based on the length dimension: —3 a — b-\-c+d+e = —1 

Based on the time dimension: —b — c — e = — 2 

a = 1 - b 
c = 2 — b — e 
d = —b 


and 

R = Cp 1 - h n b V i - h -'D~ h Vs = CpV 2 " 

Therefore 


R = f P V 2 

where f is a function of both the Reynolds number and the Mach number. 

17:4. Effect of Reynolds Number on Flow Characteristics. Equation 
(17:1) is based on the assumption that the flow in the pipe or duct is 
laminar , i.e., that the particles of fluid follow straight paths parallel with 
the axis of the channel, with the greatest velocity at this axis and zero 
velocity for a thin layer immediately adjacent to the wall of the passage; 
this, it will be remembered, is also called viscous, or streamline, flow. 
The equation indicates that, for this type of flow, the resistance R is pro¬ 
portional to the velocity gradient (dV/dy) 0 . Experiment has shown that 
laminar flow is found only at low values of the Reynolds number. As the 
Reynolds number increases above a certain critical value, which depends 
on factors such as the roughness of the wall surface and the straightness of 
the channel, the flow becomes turbulent. In turbulent flow, the particles 
of fluid acquire a radial component of velocity, and eddies are formed. 
Particles of fluid from the inner laminae of flow begin to mix with the less 
rapidly traveling particles nearer the walls of the channel and, because of 
their higher velocity, increase the velocity of flow near the edges of the 
stream. 

As might be expected, the transition from laminar to turbulent flow is 
not abrupt but occurs gradually. As the velocity of flow increases, the 
point is reached where the more rapidly traveling elements near the axis 
of the channel begin to acquire a turbulent motion. As these particles, 
due to the radial component of velocity they have now added, mix with 
the particles of fluid in adjacent laminae, they give turbulence to a wider 
section of the flow. Turbulence spreads in this manner, as the velocity 



416 


BASIC ENGINEERING THERMODYNAMICS 


of flow further increases, until the entire flow, with the exception of a very 
thin stationary layer at the wall of the duct, becomes turbulent. This 
interval during which turbulence is spreading is called the transition 
period. After the flow has become fully turbulent, the resistance varies 
more nearly in proportion to the square of fluid velocity than to its first 
power, as in laminar flow. 

In laminar flow the distribution of velocity across the pipe is parabolic, 
varying from a maximum at the axis of flow to zero at the walls; the 
velocity profile is shown as curve 1 of Fig. 17:2. Curves 2 and 3 of that 
figure show the idealized velocity profiles during the transition period and 
when full turbulence has been attained, respectively. 

When the velocity profile is parabolic, as in laminar flow, it may be 
shown that the average velocity of the fluid is one-half the maximum 


l~ L a minor flo w 
2" Tro n sit ion 
J- Turbulent flow 


N — Pipe surface 

Fig. 17:2. Velocity profiles—laminar to turbulent flow. 

velocity at the axis of the pipe. As turbulence begins to make its appear¬ 
ance, that ratio increases rapidly, as may be judged from comparison of 
the velocity profiles of Fig. 17:2, reaching a value of about 0.8 as full 
turbulence is attained. Further increase of fluid velocity still further 
increases the ratio until, for velocities of the order of those existing in 
nozzles, it becomes very nearly 1. 

The mass rate of flow through a pipe or duct is, of course, based on the 
average velocity of the fluid; in expressing the kinetic-energy term of Eq. 
(3:5), the average velocity would therefore be used. Since the kinetic 
energy varies with the square of the velocity, this introduces an error into 
its calculation; it may be shown that the kinetic energy for laminar flow 
in a pipe is V Ay 2 /g per pound and thus is twice the value as based on the 
assumption of a uniform velocity. However, as will be shown later, 
laminar flow occurs only at very low velocities, and the error is small in 
amount as compared with other terms of Eq. (3:5). When, as in nozzle 
flow, the kinetic-energy term is of major importance, we find that F av is 
very nearly equal to F max and the error in assuming a uniform velocity in 
the calculation of the kinetic energy is again small. 

As has been stated above, the point at which the flow begins to lose its 
laminar character will depend upon factors such as the roughness of the 
wall surface and the straightness of the passage. The critical Re varies 





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THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 417 


from about 1200 to around 2000. Fully turbulent flow may be reached 
at Reynolds numbers in the usual range of 2500 to 4000. Even the upper 
value of 2000 for the critical Reynolds number is very low; it throws 
nearly all of the flows with which the engineer is concerned in practice 
into the range of full turbulence. For example, in the flow of water at 
100°F through a 1-in. pipe, it corresponds to a velocity of about 0.17 fps; 
in the flow of air at 14.7 psia, 70°F, through a round duct of 1 ft diameter, 
it is equivalent to a velocity of approximately 0.33 fps. The chief interest 
of the engineer in the resistance to fluid flow is centered in the pressure 
drop that must be maintained to balance it and in the power requirement 
that results from that pressure drop; at velocities of this order, pressure 
drops would be so small as, in his estimation, to be negligible. One of the 
fields of engineering interest in which laminar flow may be of importance 
is found in situations where the pressure to propel the fluid must be 
furnished by convective effects, i.e., by differences in density of the fluid 
itself; here the available heads, and thus the velocities, are low. An 
example is the gravity-flow hot-water heating system. 

The resistance in laminar flow is proportional to the first power of the 
velocity. Therefore, if Eq. (17:3) is to be used to calculate the resistance, 
the friction coefficient f should vary inversely with the velocity, i.e., 
inversely with the Reynolds number in the flow of the same fluid through 
the same pipe. Experiment has shown that, for laminar flow in pipes and 
ducts, 



(17:4) 


and that this value is independent of a reasonable variation in the rough¬ 
ness of the pipe surface. 

For fully turbulent flow in smooth tubes, various experimentally deter¬ 
mined expressions for f are available, all of which give its value in terms 
of the Reynolds number and all of which are in fairly close agreement. 
We shall arbitrarily select from these the relation 

f = 0.0395(Re)-** (17:5) 

* This is the Blasius relation and is selected because of its simplicity. A formula 
of more general acceptance is the Nikuradse equation: 

f = 0.0004 + 0.0276(Re)“ 0 - 237 

This equation also applies to turbulent flow in smooth tubes. For commercial pipe 
in small sizes (about 1 in.), values of f are about 30 per cent higher than as obtained 
from the Nikuradse equation. The percentage addition that is made to f as obtained 
from either the Blasius or the Nikuradse equation is based on the relative roughness. 
This is expressed in terms of a roughness factor, defined as the ratio of the radius of the 
pipe to the depth of the surface projection. The percentage increase in f will be less 
for higher values of the roughness factor and is thus less for the larger sizes of com¬ 
mercial pipe. 



418 


BASIC ENGINEERING THERMODYNAMICS 


as applying to turbulent flow in smooth tubes. The resistance to turbu¬ 
lent flow is considerably increased by roughness of the wall surface, and 
the use of Eq. (17:5) is correspondingly limited, but it will serve our pur¬ 
poses in thermodynamic analysis. Equation (17:5), in conjunction with 
Eq. (17:3), shows a variation of resistance in proportion to somewhat less 
than the second power of the velocity. 

In Fig. 17:3, Eq. (17:4) is plotted in the range of Reynolds numbers 
that is associated with laminar flow, and Eq. (17:5) is used to show the 
variation of the friction coefficient with Reynolds number in the earlier 
phases of turbulent flow. Since the chief purpose of this figure is to show 



Reynolds number 

Fig. 17:3. Variation of friction coefficient at low Reynolds number—-smooth tubes. 

the transition between laminar and turbulent flow, only the lower Rey¬ 
nolds numbers are included. Two transition curves are shown by the 
lines numbered 1 and 2; curve 1 assumes that the critical Re is about 1200 
and that the flow is fully turbulent at Re = 2500, while curve 2 places 
these points at Re = 2000 and Re = 4000, respectively. The curves are 
dashed to indicate that the value of the friction coefficient is somewhat 
unpredictable in this transition period; experiment has indicated that it 
will depend on whether the velocity (the Reynolds number) is increasing 
or decreasing at the moment. In fact, it may be found that the transition 
will be similar to that shown as curve 2 if velocities are gradually increased, 
while, for decreasing velocities, the return to laminar flow will approxi¬ 
mate curve 1. 

Example 17:44. Assuming the critical value of the Reynolds number to be 2000, 
calculate the highest average velocity for laminar flow when (a) water at 100°F flows 
through a 1-in. pipe and (6) air at 14.7 psia, 70°F, flows through a round duct of 1 ft 
diameter. What is the maximum velocity in each case? 









THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 419 


Solution: 


(a) The inside diameter of a 1-in. pipe is 1.05'in. 


p = = 1.93 lb-sec 2 /ft 4 

n = 2.317 — (0.0164) (100) = 0.677 centipoise 



or 


F av = 0.168 fps 
F max = 2 Fav = 0.336 fpS 



= 0.00232 lb-sec 2 /ft 4 


n = 0.0165 + (2.5 X 10 5 ) (70) = 0.0183 centipoise 


VD P _ V (1) (0.00232) (47,800) 


0.0183 


or, 

Fav = 0.33 fps 

Fmax = 2 Fav = 0.66 fpS 

Example ll'AB. Calculate the resistance corresponding to the conditions of parts 
a and b of Example 17:4A. 

Solution: 

(a) f = mo = 0.004; R = f P F 2 = (0.004)(1.93)(0.168) 2 = 0.000218 fps 
(5) f = 0.004; R = (0.004)(0.00232)(0.33) 2 = 0.00000101 psf. 

Example 17:4C. Find the resistance and the maximum velocity when the average 
velocity is 30 fps in the flow of (a) water at 100°F through a smooth tube of 1 in. inside 
diameter and ( b) air at 14.7 psia, 70°F, through a smooth duct of 1 ft inside diameter. 


Solution: 


(30) (1) (1.93) (47,800) 
(12) (0.677) 


= 340,000 


This is turbulent flow, and Eq. (17:5) applies 


f = (0.0395) (340,000)-i = = 0.00164 

R = f p F 2 = (0.00164)(1.93)(30) 2 = 2.84 psf 


Tr _ T av _ 30 _ _ - 

_ 0.8 _ 0.8 “ 37 ' 5 fp 


,,, „ _ (30) ( 1) (0.00232) (47,800) 

(h ' Re 0.0183 


= 182,000 (turbulent flow) 


f = (0.0395) (182,000)“! = ^5? = 0.00192 

R = (0.00192)(0.00232)(30) 2 = 0.004 psf 
V max = 37.5 fpS 















420 


BASIC ENGINEERING THERMODYNAMICS 


17:5. Pressure Drop in Pipes and Ducts. In the design of pipe and 
duct systems, the primary concern of the engineer is to select a size of 
conduit which, taking into consideration the length of the channel and 
the pressure differential that is available, will deliver the fluid at the 
desired rate. The calculation of the pressure drop that corresponds to a 
given rate of flow is therefore a matter of principal importance to him. 

The pressure drop must be large enough to overcome the resistance to 
flow offered by friction between the fluid and the walls of the passage. 
In addition, if there has been a significant increase in specific volume and 
thus in the velocity of the fluid during its flow through the pipe, the pres¬ 
sure drop must also account for the increased momentum of the fluid. 
The change of specific volume is primarily due to the lower pressure of the 
fluid that has resulted from the pressure drop necessary to maintain flow 
at the required rate. If the fluid is a liquid, or if a gas and the pressure 
drop is only a small fraction of the pressure at entrance to the line, the 
effect of increased momentum is small and may be ignored. This is the 
simplified form of the problem with which the engineer usually deals and 
will be the subject of our initial discussion. 

Considering a round pipe of length L, the total force that opposes flow 
is the resistance multiplied by the wall area, or R/rDL. This force is over¬ 
come by the pressure drop acting on the inside cross-sectional area of the 
pipe, or AP(7rD 2 /4). Therefore we may write 


RttDL = A P 


ttD 2 

4 


or 


4PL __ 4fpF 2 L 
D ~ D 


(17:6) 


For a conduit of other than circular cross section we may write 


or 


P(circumference)L = AP(cross-sectional area) 


A P 


RL 

r 


f P V 2 L 
r 


(17:7) 


in which r is the hydraulic radius, or (cross-sectional area)/(inner circum¬ 
ference). Equations (17:6) and (17:7) will apply to the calculation of the 
pressure drop in gas flow with sufficient accuracy for the usual engineering 
purpose if the total pressure drop does not exceed one-tenth of the absolute 
pressure at entrance to the line. Average values of f, p, and V are used. 

The value of f in Eqs. (17:6) and (17:7) will be based on the Reynolds 
number. When the conduit is not circular, 4 r may be substituted for D 
in computing the Reynolds number (Re = 4rFp/p) if the flow is turbu- 







THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 421 


lent, as is usually the case when the pressure drop is large enough to be 
important. 

Example 1/ :5. Find the pressure drop per 100 ft length of line (a) when water at 
100°F flows in a 1-in.-inside-diameter smooth tube with an average velocity of 30 fps 
and (6) when air initially at 14.7 psia, 70°F, flows through a smooth round duct of 1 ft 
diameter with an average velocity of 30 fps. 

Solution: 


(a) From Example 17:4C, f =0.00164; p = 1.93; D = 

\ p _ 4fp V 2 L (4)(0.00164)(1.93)(30) 2 (100)(12) _ . 

— — j )— =-^- = 13,600 psf = 94 psi 


(6) From Example 17:4C, f = 0.00192; p = 0.00232. 


AP 


4f P V 2 L _ (4)(0.00192)(0.00232)(30) 2 (100) 
D 1 


= 1.6 psf = 0.011 psi 


When the fluid is a gas and the pressure drop is large, it becomes neces¬ 
sary to divide the entire length of pipe into shorter sections in order that 
the effect of change of velocity on the pressure differential that is required 
may be studied. In the limit, this method of dividing the passage into 
shorter lengths would arrive at a length dL for which the pressure change 
dP is to be computed. In passing through this length the velocity of the 
fluid increases by dV and the momentum by M dV/g. Applying Eq. 
(3:8) to a pipe of constant cross-sectional area, V = Mv/A or, since M 
and A are both constant, dV = (M dv/A). The force acting on the 
cylindrical element of fluid by reason of the pressure change dP is A dP, 
and the frictional force dF applied at the wall surface is fpTVD dL. The 
sum of these three forces must total zero if the pipe is horizontal, and we 
may write 

M dV 

AdP + + f pVhrD dL = 0 (a) 

Dividing each term by Av and substituting (M/A) dv for dV and Mv/A 
for V, this becomes 

dP 1 /mV dv 4i(M/AydL = 0 

v g\A) v gD 


Integrating between sections 1 and 2, which are a distance L apart, we 
have 



1 I 

l0g e -b 

Vl 


4f(M/A) 2 L 

gD 


= o 


(17:8) 


in which M/A is the weight rate of fluid flow per unit of cross-sectional 
area. To evaluate the first two terms of this expression, a knowledge of 










422 


BASIC ENGINEERING THERMODYNAMICS 


the relation between the pressure and the volume of the fluid in passage 
through the pipe is required. Applying Eq. (3:5) to adiabatic flow 
through a horizontal pipe, 


hi + 


TV 

2 Jg 


/&2 ~b 


n 

2 Jg 



or 

h + m = h + w g (5) **- const = h ° (17:9) 

in which ho is the value approached by h as M/A approaches zero and may 
be readily computed based on the state of the fluid at any section of the 
pipe when M/A has a known value. 



17 :6. Fanno Lines. When Eq. (17:9) is plotted on a Mollier chart for 
a given entry state 1 and flow rate M/A, as in Fig. 17:4, the resulting 
curve is called a Fanno line. 1 Points along this line indicate the state of 
the fluid resulting from adiabatic flow at the rate M/A as the pressure in 
the pipe drops from the entry pressure p i to the successively lower pres¬ 
sures P 2 , Vh Vh an d Pb- At point 3 the path becomes vertical (isentropic) 
and, below that point, curves back to lower entropies. But the process is 
adiabatic and, according to the Second Law [see (Eq. 6:7)], the entropy 

1 A. Stodola, “Steam and Gas Turbines,” Vol. 1, p. 61, McGraw-Hill Book Com¬ 
pany, Inc., New York, 1927. 








THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 423 


cannot decrease. Therefore the expansion cannot proceed beyond point 
3 within the pipe. If a lower pressure than p 3 is maintained in the space 
into which the pipe discharges the fluid, the pressure drop from p 3 to this 
lower pressure must take place after the pipe exit has been passed, since 
it will require a larger area for fluid flow than the pipe provides. Maxi¬ 
mum flow through the pipe is attained as the pressure in the exhaust space 
becomes p 3 ; any further decrease of pressure in this space will not change 
the pressure drop in the pipe and so will not increase the rate of flow 
through it. 

Let us suppose that maximum flow is obtained by reducing the pressure 
at exit from the pipe to p 3 ; at point 3, ds = 0, 


j . Pv „ , . Pdv . vdP m ,.vdP vdP , , . , 

h — u ~\~ —j- dh = du H— ~j —1 —~j - I ds T — -j — — —j — (at point 3) 


Differentiating Eq. (17:9), 


„ , i (mV , vdP , i (mV 

dh + Tg{-A) v dv = ~j ^ Jg ("A / vdv 


= 0 


or 


and 


From Eq. (8:25), 


and, substituting, 


dv\ 1 _ v 2 

~ ~ 9 (M/A) 2 ~ ~ 9 V 2 


dP 


dP\ = h (dP\ = -kP z v , 


dv 


dv J t 


v i 


F 3 = NkgP?Vz 


But this is the velocity of sound in the gas at state 3, and therefore F 3 = 
V s . When the gas enters the pipe at a velocity below the velocity of 
sound, the acoustic velocity is the highest velocity that can be attained 
within the pipe. 

Points below 3 on the Fanno line of Fig. 17:4 correspond to velocities 
greater than the velocity of sound and cannot be reached through point 
3 without allowing the gas to escape from the pipe. However, if the gas 
enters the pipe at a velocity greater than that of sound (as at 4 or 5), it 
will pass through the series of states that are connected by the Fanno line 
but in the reverse direction (toward increased pressure and entropy). 
Thus friction in the pipe will bring about a rise in pressure if the velocity 
exceeds the acoustic velocity. Point 3 is again the limit of this process 
since beyond that point the entropy decreases. If the pressure in the 












424 


BASIC ENGINEERING THERMODYNAMICS 


space into which the pipe discharges is higher than p 3 , the last stage of the 
compression must take place after the gas has left the pipe. 

Example 17:64. Plot the Fanno line for the flow of steam with M/A = 300 
lb/(sec)(ft 2 ). At section 1, the pressure is 400 psia and the temperature 600°F. 
Solution: 


pi = 400 psia; h = 600°F; Vi = 1.4770 ft 3 ; hi = 1306.9 Btu; Si = 1.5894 

Vi = ~ vi = (300) (1.4770) = 443 fps 
A 


h 0 = hi + 



= 1306.9 + 


(300) 2 (1.4770) 2 

50,000 


1310.8 Btu 


Let p 2 = 200 psia. The pressure has been halved, and, for a trial, we shall assume 
that the specific volume, and therefore the velocity, has been doubled. Then the trial 

value of h 2 is h 0 - = 1310.8 - (3QQ)2( | q 000 ~' ^ = 1295 - X Btu ‘ At 200 psia ’ 

this enthalpy lies between temperatures of 540 and 560°F. Making trial calculations 
at each of these temperatures, for 540°F 


h 0 = 1290.5 + 
and for 560°F 
ho = 1301.1 + 


(300) 2 (2.862) 2 

50,000 

(300) 2 (2.928) 2 

50,000 


= 1305.3 Btu 


1316.5 Btu 


Interpolating for the required value of h 0 = 1310.8, the temperature t 2 is found to be 
about 550°F, and the other properties are v 2 = 2.895; h 2 = 1295.8; s 2 = 1.6513. 
Checking, 

h 0 = 1295.8 + (1.8)(2.895)2 = 1310.8 

This locates a second point on the Fanno line. The velocity V 2 = (M/A)v 2 = 
(300) (2.895) = 870 fps. Proceeding in like manner, the following table may be pre¬ 
pared from which the Fanno line may be plotted, as in Fig. 17:4. The pressures of 
100 and 80 psia have been added to the table to show more clearly the point at which 
the Fanno line becomes vertical. Checking at point 3 (90 psia) for V s ■ 

V 8 = VkgPzv a = \/(1.3) (32.2) (90) (144) (5.806) = 1775 fps 

This is close to the velocity V 3 but indicates that the point at which the Fanno line 
becomes vertical lies at a pressure slightly below 90 psia. 


Data for Fanno Line 


M/A = 300; pi = 400 psia; h = 600°F 


Station 

Pressure, 

psia 

Tempera¬ 
ture, °F 

Enthalpy, 
Btu 

Volume, 

ft 3 /lb 

Entropy 

Velocity, 

fps 

1 

400 

600 

1306.9 

1.4770 

1.5894 

443 

2 

200 

550 

1295.8 

2.895 

1.6513 

870 


100 

461 

1259.3 

5.339 

1.6874 

1600 

3 

90 

441 

1250.3 

5.806 

1.6889 

1745 = F s approx 


80 

415 

1238.3 

6.342 

1.6879 

1905 

4 

70 

382.5 

1223.2 

6.974 

1.6848 

2090 

5 

50 

287.5 

1177.6 

8.603 

1.6632 

2580 

























THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 425 


Ihe relation between the pressure and the volume that must be known 
it Eq. (17:8) is to be of use is based on the Fanno-line relation [Eq. (17:9)]. 
It has been shown (Example 2:9) that it is possible to express the enthalpy 
of a perfect gas in the form 


h = A + EPv (17:10) 

in which A and E are constants. 1 This relation may also be applied with 
fair accuracy to a wider range of gases and vapors, provided that the 
values of the constants are established from data at states close to those 
in the range over which the equation is to be applied. Making this sub¬ 
stitution in Eq. (17:9), we have 

A + EPv + irg(^y v2 = h ° (17:11) 

This is a quadratic equation of the form 


av 2 + EPv + c = 0 


1 (M\ 


(17:12) 


where a = \ ~[J an d c — A — h 0 . All quantities in Eq. (17:12) 

are constants except v and P. Solving for P in terms of v, 


P = - 
dP = - 


a cl 
E V ~ E~v 
a 7 . c dv 

E dv + EV> 


Substituting this value in the first term of Eq. (17:8) and integrating, 



a [ V2 dL.c_ f V2 dv 

E J Vl v ' E J Vl v* 


a 

E 



( 1 

E \v 2 2 



(17:13) 


When this value is substituted for the first term of Eq. (17:8), that equa¬ 
tion may be used for the computation of the coefficient of friction in the 
flow of compressible fluids; measurements of the mass rate of flow and 
data with respect to the state at entrance and the pressure at exit from a 
section of the pipe of measured length will be sufficient to enter the equa¬ 
tion. In obtaining the values of v\ and v 2 for substitution, we may return 
to the quadratic [Eq. (17:12)] and, solving it, obtain 


v 


-EP ± \/E 2 P 2 - 4ac 
2 a 


(17:14) 


Example 17:65. Steam is supplied to a length of 1-in. commercial pipe from a tank. 
The pressure in the tank is steadily maintained at 40 psia, the temperature is 300°F, 

1 A as in Eq. (1:7); E = ^ + j with B as in Eq. (1:7). 








426 


BASIC ENGINEERING THERMODYNAMICS 


and the steam velocity is negligible. The pressure at an upstream pipe section 1 is 
35 psia and, at a second section located 36 ft downstream from section 1, is 25 psia. 
The weight of steam flow through the pipe is 10 lb /min. The pipe is well insulated, 
and the flow may be considered to be adiabatic. Calculate the value of the friction 
coefficient, and compare with that obtained from Eq. (17:5). 

Solution. Commercial 1-in. pipe has an internal diameter of 1.05 in. and an internal 

transverse area of 0.006 ft 2 . 


M_ 

A 


10 

(60) (0.006) 


27.75 lb/(sec)(ft 2 ); h 0 = h at 40 psia, 300°F 


1186.8 Btu 


Using the method outlined in Example 17‘.64, the following data may be calculated. 

Vl = 35; tx = 292.7°F; h x = 1184.4; vx = 12.525; 8x = 1.7105; Vi = 347 fps 
p 2 = 25; t 2 = 283.1°F; h 2 = 1182.2; v 2 = 17.410; s 2 = 1.7439; V 2 = 483 fps 

Using these data to determine A and E in Eq. (17:10), 


1184.4 = A + E(35) (144) (12.525) 
1182.2 = A 4- E(25)( 144) (17.410) 


from which 


E = 0.004 and A = 931.7 
27.75 2 


a = 


= 0.0154; c = 931.7 - 1186.8 = -255.1 


50,000 

Substituting in Eq. (17:8), 

0.0154 17.410 1 /-255.1W 1 1 \ 

l0ge 12.525 2 V 0.004 ) \17.410 2 12.525 2 / 


0.004 


+ 




( 2 7.75>. log. iui?+ mmmm = 0 


12.525 


or 


1.27 - 96.5 + 7.9 + 39,400f = 0 and f = 0.00228 
n = 3 X 10 -7 lb-sec/ft 2 


(32.2)(1.05) 


[Table 6, steam tables] 




= 0.00207 


vg_ (15) (32.2) 

Re = = (415) (1.05) (0.00207) X 10- . 250>000 

n ( 12 )( 3 ) 

f = (0.0395) (250,000)"* = 0.00177 


[Eq. (17:5)] 


It will be noted that this calculation has been based on the use of average values of the 
specific volume, the viscosity, and the velocity. The value of f that is obtained from 
the assigned data is about 30 per cent higher than the value obtained from Eq. (17:5), 
which applies to smooth tubes. This is a reasonable allowance for the roughness 
factor of 1-in. commercial pipe. 


17 :7. The Rayleigh Line. Pressure shock (see Art. 11:7) is the name 
given a sudden rise in pressure which is sometimes observed in the diverg¬ 
ing section of a nozzle. Observation indicates that the fluid always enters 
this compression wave at a velocity greater, and leaves it at a velocity 
less, than that of sound and that, for steadily maintained supply condi¬ 
tions and exhaust pressure, the wave remains stationary in the channel. 














THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 427 


The pressure rise takes place so sharply that the divergence of the 
channel may be ignored and the flow may be compared with that in a very 
short length of pipe. Because the channel section is so short, wall friction 
may also be neglected and Eq. (a) of Art. 17:5 may be written as 

M - 

AdP + — dV = 0 

g 

Designating the entry section as a and the exit section from the shock 
wave as b , integration gives 

A(P b -Pa) +~ (Vb -V a )= 0 

c/ 

or, since V = ( M/A)v , 

Pb = Pa - - (Aj (Vb - Va) (17:15) 

Beginning with the known values of P and v at a and assuming successively 
lower values of Vb, this relation may be plotted on a Moilier chart, as in 



Fig. 17:5; it is represented by the curve acb of that figure and is called a 
Rayleigh line. Also shown in Fig. 17:5 is a Fanno line which has been 
drawn through state a for the same flow rate M/A as applies along the 
Rayleigh line. The two curves again intersect at state b. 




428 


BASIC ENGINEERING THERMODYNAMICS 


The fluid cannot progress through the series of states that are connected 
by the Rayleigh line without heat flow; yet the flow through the channel 
is adiabatic. However, at state b the Rayleigh line returns to intersect 
again the Fanno line, which connects states assumed in adiabatic flow. 
State b is consequently not inconsistent with the requirement that the 
flow be adiabatic; it is therefore a possible state to which the fluid may 
suddenly shift. The pressure at b will always be higher than at point a; 
the same is true of the entropy. The sudden change from state a to state 
b is possible only because the entropy has increased during this adiabatic 
process; a change in the opposite direction (from b to a) would, according 
to the Second Law, be inconceivable, and this explains why the pressure 
is always observed to rise, never to fall, in these shock waves. 

To find the second point of intersection of the Rayleigh and Fanno 
lines at state b, we may assume as before that Eq. (17:10) provides a 
sufficiently accurate relation and write 


hb — A + EPbVb 


(17:16) 


Substituting this value of h b in the equation of the Fanno line [Eq. (17:9)] 
we have 

A + EP b v b + A- (T) v b - = ho 

or 


Pb = 


ho — A 


i /mV 

VgE\AJ Vh - 


c 1 
E v h 


a 


Ev b 2JgE 
Equating this value of P b to that of Eq. (17:15), 

2 Jav b + 2 Jav a 


E 


v b 


(17:17) 


c 1 a 
E V b ~ E Vb ~ Pa 


(17:18) 


Multiplying by Ev b , rearranging, and changing signs, 

(a — 2 JaE)vb 2 + (EP a + 2 ,JaEv a )vb + c = 0 (17:19) 

This is the quadratic form and can be solved for v b . Values of P b and h b 
may be computed by substitution in Eqs. (17:17) and (17:16) in that 
order. 


Example 17:7. Plot a Rayleigh line for M/A = 300 lb/(sec)(ft 2 ) which passes 
through state 5 of the Fanno line of Example 17:6 A. Find the second intersection 
with the Fanno line of that example. 

Solution: 

P a = P- 0 = (50) (144) = 7200 psf; v a = v 5 = 8.603 ft 3 /lb 

A series of points on the Rayleigh line will be based on arbitrarily assumed increasing 
pressures until the second point of intersection with the Fanno line is reached. To 
check the location of that point, Eq. (17:19) will be solved for v h . Increasing pres¬ 
sures are assumed instead of decreasing volumes since they offer a more convenient 






THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 429 


key with which to enter the steam tables for interpolation. A sample calculation is 
shown below. 

Assume p b = 70 psia. Substituting in Eq. (17:15), 

Qf>02 

(70) (144) = (50) (144) - ^ (v b - 8.603) or = 7.573 ft 3 /lb 

Taking a pressure of 70 psia and specific volume of 7.573 ft 3 /lb to Table 3 of the 
steam tables, interpolation gives the temperature as 448°F, the enthalpy as 1256.2 Btu, 
and the entropy as 1.7224. Following this method for successively higher pressures, 
the following table is prepared: 

Data for Rayleigh Line 


M/A = 300; Va = 50; v a = 8.603 


Station 

Pressure, 

psia 

Volume, 

ft 3 /lb 

Temp., 

°F 

Enthalpy, 

Btu 

Entropy 

a 

50 

8.603 

287.5 

1177.6 

1.6632 


60 

8.088 

375 

1221.0 

1.6986 


70 

7.573 

448 

1256.2 

1.7224 


80 

7.058 

505 

1283.5 

1.7371 


90 

6.548 

546 

1303.9 

1.7443 


100 

6.033 

570 

1314.4 

1.7439 


110 

5.518 

578 

1317.4 

1.7366 


120 

5.003 

569 

1312.2 

1.7221 


130 

4.483 

543 

1298.2 

1.6997 


140 

3.973 

504 

1277.3 

1.6705 

b 

139 

4.04 

512 

1281.4 

1.6756 


The data of the second to the ninth lines of the table have been obtained by the 
method of calculation outlined above. The data have been used to plot points on 
the Mollier chart of Fig. 17:5, and these points have been connected to form the 
required Rayleigh line. It was found that this line recrossed the Fanno line between 
pressures of 130 and 140 psia. To check the point of intersection, resort was made 
to Eq. (17:19). To obtain A and E for use in that equation, the two states selected 
were state a and the state, lying near the second intersection, where p = 140 psia, 
v = 3.973 (see table above). Then, applying Eq. (17:10), 

1177.6 = A + E(50) (144) (8.603) 

1277.3 = A + E(140)(144) (3.973) 

and, solving, 

E = 0.0055; A = 837.0 


Also, 
a = 


300 2 

50,000 


1.8; c = 837.0 - 1310.8 


-473.8 


a + 2 JaE = 1.8 - (2) (778) (1.8) (0.0055) = -13.62 

EP a + 2 JaEva = (0.0055) (50) (144) + (2) (778) (1.8) (0.0055) (8.603) 


172.1 


Changing signs, Eq. (17:19) becomes 
13.62v 6 2 - 172 Av b + 473.8 = 0 


















430 


BASIC ENGINEERING THERMODYNAMICS 


or, solving for v b , 


172.1 ± V172.1 2 - (4) (13.62) (473.8) 
(2)(13.62) 


4.04 


This volume has been used in calculating the data for the last line of the table, giving 
the properties at the second point of intersection. The pressure at entrance to the 
wave defined by this Fanno-Rayleigh line combination is 50 psia, and the velocity 
(see table of Example 17:64) is 2580 fps, well above the velocity of sound. At exit 
from the wave, the pressure has risen sharply to 139 psia, and the velocity is 


V b = TI Vb = (300) (4.04) = 1212 fps 
A 


This velocity is well below the velocity of sound. It will also be noted that s b > s a . 


Problems 

1. Show that the unit of viscosity used in Table 6 of the Keenan and Keyes steam 
tables (lb-sec/ft 2 ) is equal to 47,800 centipoises. 

2. A thin rectangular plate 50 by 70 in. is immersed in a stream of lubricating oil 
(SAE 30). What is the viscous drag on the plate in pounds if the velocity gradient 
at the surface of the plate is 5 fps/in.? Assume t = 70°F. 

3. The viscous drag forms a part of the total drag of an airplane wing and is often 
called the “skin-friction” drag. It is calculated from the following equation: 

D, = C,t SV* 


in which D/ = skin-friction drag, lb 

Cf = dimensionless coefficient 
p = mass density of air, slugs/ft 3 

S = projected area of wing (area of one side, approximately), ft 2 
V = air speed relative to wing, fps 

If C/ = 0.006, and the wing is rectangular, 30 ft from tip to tip and 5 ft from leading to 
trailing edge, what is the skin-friction drag of the wing in flying through air at standard 
atmospheric pressure and a temperature of 59°F at a speed of 100 mph? Based on 
Eq. (17:1), find the velocity gradient at the surface of the wing. 

4. Employ dimensional analysis to show that the dimensionless coefficient Cf in 
the formula presented in Prob. 3 for the calculation of drag due to skin friction is a 
function of the Froude number ( V 2 /Lg ). Assume that the factors affecting the drag 
are a dimension of the wing (L), the velocity of the air stream (P), the mass density 
of the air (p), and the acceleration due to gravity ( g ). 

5. Show that if the velocity of sound is added to the list of factors affecting the 
resistance to flow [see Eq. (1), Art. 17:3], the coefficient of friction becomes a function 
of the Mach number as well as the Reynolds number. 

6. Prove that the kinetic energy for laminar flow in a pipe, assuming a parabolic 
velocity profile, is P av 2 /</ per pound. 

7. What average velocity is equivalent to Re = 2000 in the flow of SAE 30 lubri¬ 
cating oil at a temperature of 100°F (density = 57 lb/ft 3 ) in a 1-in. pipe? What 
is the maximum velocity? Repeat for a |-in. pipe (internal diameter = 0.62in.). 

8. Check Fig. 17:3 at Reynolds numbers of 1000, 2000, 4000, and 10,000. Calcu¬ 
late the friction coefficient for a smooth tube at a Reynolds number of 100,000. 

9. Compare the values of the friction coefficients for smooth tubes as obtained 




THE STEADY FLOW OF FLUIDS IN PIPES AND DUCTS 431 


from the Blasius and the Nikuradse equations at a Reynolds number of 4000. At 
Re = 10,000. At Re = 100,000. 

10. Calculate the total drag in a 100-ft length of commercial 1-in. pipe when it 
carries water at 60°F and Re = 2000. What weight of water is handled per hour? 
Repeat for Re = 10,000. Note that this is commercial pipe. What differential of 
pressure must be maintained between the ends of the pipe in each case? 

11. Air at 130°F, 14.7 psia, moves through a smooth duct of 2 ft inside diameter 
at a Reynolds number of 200,000. The duct is 100 ft in length. Assume that the 
temperature of the air does not change in passage through the duct, (a) Calculate 
the total drag, assuming that the velocity is the same at all sections. ( b ) What 
volume, in cubic feet per minute, is handled? (c) What is the pressure drop? 
( d ) What power is required? ( e ) What percentage change in velocity would actually 
be realized if account is taken of the compressibility of the air? 

12. (a) Repeat Prob. 11 for a duct 2 ft square in section. ( b) For a rectangular 
duct 10 by 20 in. 

13. Air at 100°F, 15 psia, enters a 12- by 20-in. rectangular smooth duct 100 ft 
long. At exit, the pressure is 14.9 psia, and the temperature is 96°F. What weight 
of air is handled per minute? (Note: Assume a reasonable value for Re, and check 
the calculated velocity against your assumption, making a new estimate if necessary.) 

14. The water pressure in a street service main is 60 psia. It is desired to choose 
a size for a branch line such that it will be capable of delivering at least 4 ft 3 of water 
per minute to a residence. The branch will have an equivalent length of 100 ft. 
(The equivalent length of line allows for the resistance of fittings, etc., by expressing 
this resistance in terms of lengths of straight pipe.) Add 30 per cent to f to allow for 
roughness of commercial pipe. What size of commercial steel pipe is required? 
Refer to a handbook for the dimensions of commercial pipe. 

15. Show that the dimensions of each term of Eq. (17:8) are the same. 

16. Solve Example 17:6A when (a) M/A = 1500 lb/(sec)(ft 2 ) and ( b ) when 
M/A = 1280 lb/(sec)(ft 2 ). Plot these Fanno lines and the Fanno line of Example 
17:6A on the same graph. 

17. As M/A approaches 0, what locus is approached by the Fanno line? 

18. Steam enters a pipe with an internal diameter of 2 in. from a large tank in 
which its pressure is 100 psia, its temperature is 400°F. At a certain section of the 
pipe, the pressure has decreased to 80 psia, the temperature to 390°F. Assuming 
the flow to be adiabatic, what weight of steam per minute is carried by the pipe? 
If this flow through the pipe is a maximum, what are the pressure and temperature 
of the steam at its exit? 

19. Steam is withdrawn from a large tank in which its pressure is 100 psia and its 
temperature 400°F through a well-insulated pipe with an internal diameter of 2 in. 
The rate of steam flow is 75 lb/min. A{ a certain section of the pipe, the pressure is 
80 psia. What is the temperature at this section? At a section that is 85 ft farther 
downstream, the pressure is found to be 60 psia. What is the temperature at this 
second section? What is the coefficient of friction? 

20. Air is withdrawn from a large tank, in which its pressure is 100 psia and its 
temperature 400°F, through a well-insulated pipe with an internal diameter of 2 in. 
The rate of air flow is 75 lb/min. At a certain section of the pipe, the pressure is 
80 psia. What is the temperature at this section? At a section farther downstream, 
the pressure of the air is 60 psia. If the average value of f between the two sections 
is 0.0018, what is the distance between them? 

21. Plot a Rayleigh line for M/A = 300 lb/(sec)(ft 2 ) which passes through 
state 4 of the Fanno line of Example 17:6A. Find the second intersection with the 
Fanno line. 


432 


BASIC ENGINEERING THERMODYNAMICS 


a, b, c, d, e 
A 
B 
C 
D 
E 

f 

f 

F 

9 

h 

J 

L 

M 

V 
P 

r 

R 

Re 

s 

T 

v 

V 

V 

Vs 

y 

Greek Letters 
Q 


Symbols 

constants 

area; also, a constant 
a constant 

a dimensionless constant 

a dimension of the channel; for round pipes, the diameter 
a constant 
a function 

coefficient of friction 

force; also, the dimension of force 

acceleration of gravity 

specific enthalpy 

proportionality factor 

length; also, the dimension of length 

mass rate of flow; also, the dimension of mass 

pressure, psi 

pressure, psf; pressure in general 
hydraulic radius 

force per unit area of wetted surface; also, the gas constant 

Reynolds number 

specific entropy 

absolute temperature 

specific volume 

volume 

velocity 

velocity of sound 

thickness of a fluid lamination 


p 


the dimension of time 
coefficient of viscosity 
mass density 


CHAPTER 18 


REAL GASES 

18:1. Introduction. The fluids with which we have been concerned in 
preceding chapters may be put into two classifications. They either were 
perfect gases or were fluids, which, like steam, are important enough to 
the engineer and vary sufficiently from the behavior of a perfect gas in the 
range of states of engineering interest to warrant the extensive research 
that is required to prepare special equations of state and, from these equa¬ 
tions, tables of properties. Our purpose in restricting ourselves to these 
two classes of fluids has been to avoid complications and distractions in 
the development of basic thermodynamic theory. In line with that pur¬ 
pose, we have further simplified the handling of a perfect gas by assuming 
that its specific heats are constant although, as shown in Chap. 9, there is 
no reason why the specific heat of even a perfect gas may not be a function 
of the temperature. For this reason, equations such as (9:27) to (9:43) 
must be regarded as only approximations as they apply to the real gas, 
even when the fluid is one of the “permanent” gases. Whether these 
equations, as written, are sufficiently accurate for the engineer’s purposes 
will depend upon the degree of accuracy that is required. In addition, 
it may be necessary to deal with fluids which, in the range of states of 
interest, are close to the two-phase region where the perfect-gas relation is 
entirely inadequate; if special tables of properties are not available to his 
use, nothing in preceding chapters would equip the engineer to deal with a 
situation of this kind. Although a full treatment of the subject is beyond 
the scope of this text, it will be the purpose of the present chapter to intro¬ 
duce the reader to the methods which are applied in such circumstances. 

18:2. The Perfect Gas with Variable Specific Heat. At low pressure 
the perfect-gas relation will, for all gases, hold to a high degree of accuracy. 
Even when the pressure is a considerable fraction of the critical pressure 
but the temperature is high as compared with the critical temperature, 
its accuracy is often satisfactory for the purposes of the engineer and he 
may treat the gas as a perfect gas. 

That he considers the perfect-gas relation, Pv = RT, a sufficiently 
accurate relation to apply to the situation at hand does not, however, 
mean that the engineer may safely take the further step of assuming con¬ 
stant specific heats in simplifying his problem. Note, however, that if 
the perfect-gas relation is assumed to hold, the variation of specific heat 

433 


434 


BASIC ENGINEERING THERMODYNAMICS 


Table 18:1. Equations for Specific Heat at Constant Pressure 


Eq. 

No. 

Gas 

Mol. 

wt 

Molar specific heat at constant pres¬ 
sure, mcp, Btu/(mole)(°R) 

(T in °R) 

Range, 

°R 

Max 
error, % 

(D* 

(2) f 

(3) t 

(4) | 

o 2 

32.00 

6.40 + 0.001137 7 

6.93 + 0.0000001254 7 72 

11.515 - 172 + 15307 7-1 

11.515 - 172 T~* + 15307 7 " 1 

+ (0.00005) (T - 5000) 

720-1900 
Not stated 
540-5000 
5000-9000 

1 

Not stated 
1.1 

0.3 

(5) * 

(6) t 

(7) t 

n 2 

28.02 

6.26 + 0.0008187 7 

6.93 + 0.0000001254 T 72 

9.47 - 34707 7-1 + 1,160,000T- 2 

720-1900 
Not stated 
540-9000 

1 

Not stated 
1.7 

(8)* 

(9)t 

Air 

28.96 

6.37 + 0.000886 T 

6.93 + 0.0000001254 7 72 

720-1900 
Not stated 

1 

Not stated 

(10)* 

(H)t 

(12) t 

(13) t 

h 2 

2.016 

6.75 + 0.0002307 7 

6.93 + 0.0000001254 7 12 

5.76 + 0.0005787 7 + 207 7 "* 

5.76 + 0.0005787 7 + 207 7 "* 

- (0.00033) (T - 4000) 

720-1900 
Not stated 
540-4000 
4000-9000 

1 

Not stated 
0.8 

1.4 

(14) * 

(15) f 

(16) t 

(17) } 

C0 2 

44.00 

8.18 + 0.002757 7 

7.15 + 0.00397 7 - 0.00000067 72 
12.196 + 0.000427 1 

16.2 - 6530 7 7-1 + 1,410,000T 7-2 

720-1900 
Below 2900 
Above 2900 
540-6300 

3 

Not stated 
Not stated 
0.8 

(18)* 

(19) t 

(20) J 

CO 

28.00 

6.33 + 0.0008987 7 

6.93 + 0.0000001254 7 12 

9.46 - 3290 7 7-1 + 1,070,0007 7 - 2 

720-1900 
Not stated 
540-9000 

1 

Not stated 
1.1 

(21)* 

(22) f 

(23) t 

H 2 0 

18.016 

7.80 + 0.0002997 7 

7.80 + 0.00037 7 + 0.0000003157 72 
19.86 - 597 T~* + 75007 7 - 1 

720-1900 
Not stated 
540-5400 

Not stated 
Not stated 
1.8 

(24) * 

(25) f 

(26) t 

ch 4 

16.03 

3.33 + 0.00899 T 7 

3.459 + 0.010567 7 

4.52 + 0.007377 1 

720-1900 
Not stated 
540-1500 

Not stated 
Not stated 
1.2 

(27)| 

c 2 h 4 

28.03 

4.23 + 0.011777 7 

350-1100 

1.5 

(28) t 

c 2 h 6 

30.05 

4.01 + 0.016367 7 

400-1100 

1.5 

(29) t 

c 8 h 16 

112.3 

7.92 + 0.0601 T 

400-1100 

4 

(30)* 

c 8 h 18 

114.1 

11.98 + 0.005557 7 

720-1900 

Not stated 


* Adapted from NACA Rept. 699 (1940). 

t Marks, “ Mechanical Engineers’ Handbook,” 5th ed., McGraw-Hill Book Company, Inc., New York, 
1951. 

X Sweigert and Beardsley, ‘‘Empirical Specific Heat Equations Based on Spectroscopic Data” (at 
zero pressure), Georgia School Technol. State Eng. Expt. Sta. Bull. 1, No. 3 (1938). 


























































REAL GASES 


435 


is only with temperature and is independent of the pressure. Moreover, 
the difference, at any given state, between the specific heat at constant 
pressure and the specific heat at constant volume must always equal R, 
itself a constant. 

Equations for the specific heats of real gases are based on experimental 
data obtained by various methods of approach. Being experimentally 
derived, they must be regarded as only closer approximations than the 
use of a constant value for the specific heat. The specific heat of the 
multiatomic gas that follows the perfect-gas relation has been shown in 
Chap. 9 to increase with temperature. Empirical equations that express 
its value must reflect this requirement if they are to be considered con¬ 
sistent. These equations are in-various forms; three typical forms are 
listed below, 

c p = A + BT + CT 2 ( 1 ) 

c p = A + BT-* + CT~' + D{T - E) (2) 

Cp = A + BT-' + CT- 2 (3) 

in which A, B, C, D, and E are constants. Of these constants, B, C, D, 
and E may have zero values in a specific equation or may be negative. 
For example, it will be noted that if the specific heat is to show the charac¬ 
teristic increase with temperature, either B or C, or both, must be negative 
in (3). An equation of the form of (1), but with C = 0, is usually satis¬ 
factory to yield the degree of accuracy required in engineering calcula¬ 
tions. Table 18:1 lists a number of equations from various sources from 
which a choice may be made according to the degree of accuracy desired. 
The equations based on spectroscopic data cover a wider temperature 
range and are credited with being the most precise; they are also, in 
general, the most complicated from the standpoint of ease of use in 
establishing thermodynamic relations between primary and secondary 
properties. All of the tabulated equations give the molar specific heat 
at constant pressure [Btu/(mole)(°R)]; to obtain the specific heat per 
pound, each constant must be divided by the molecular weight of the gas. 
The equation for the specific heat at constant volume may be established 
from these expressions by deducting the quotient obtained by dividing 
the universal gas constant by J, or mR/J = -rrf- = 1.985, from the 
constant term (A). Thus, from Eq. (1) of Table 18:1, the equation for 
the molar specific heat at constant volume of 0 2 may be written as 

mc v [Btu/ (mole)(°R)] = (6.40 - 1.985) + 0.00113T = 4.415 + 0.001137 7 

The introduction of a variable specific heat which is a function of the 
temperature alone brings about certain changes in the equations for the 
heat flow and work that accompany a perfect-gas reversible process and 
the expressions which define the change of internal energy, enthalpy, and 
entropy. 



436 


BASIC ENGINEERING THERMODYNAMICS 


Changes of Enthalpy and Internal Energy. It has been shown [Eq. 
(9:13)] that dh = c p dT for the perfect gas. When c p is constant, 
h 2 — hi = c p (T 2 — T i). Remembering that c p is now, though variable, 
a function of T alone, we may write 

'A s -fci= jf CpdT (18:1) 

The indicated integration cannot, of course, be performed until an expres¬ 
sion which defines c p in terms of T is available. Similarly, based on Eq. 
(9:12), 

u 2 — U\ = f c v dT (18:2) 

JT i 

Changes of Entropy. The changes of enthalpy and of internal energy 
are functions of the temperature alone, but the change of entropy is a 
function of pressure as well as temperature. It has been shown in Art. 
8:1 that T ds = dh — v dP/J. For the perfect gas this may be changed 
to the form 

Tds = c p dT - 
or 


Integrating, 

S 2 — Si 

Work and Heat Flow for the Reversible Process. For our purpose, 
reversible processes will be classified as constant-pressure, constant- 
volume, isothermal, adiabatic, and polytropic, as in Chap. 9 and Table 
9:1. Each is discussed separately below: 

1. Constant-pressure (n = 0). The final pressure, temperature, and 
volume may be found from the perfect-gas relation, and the expressions in 
the third, fourth, and fifth columns of Table 9:2 may be used. The heat 
supplied the system is equal to the change of enthalpy, or 

1 Q 2 = M ff <•„ dT 

The work is, as in Table 9:2, iW 2 = P(V 2 — Fi). 

2. Constant volume (n = « ). Table 9:2 may be used to find the final 
pressure, temperature, and volume. The heat flow is equal to the change 
of internal energy, or 

1 Q 2 = M J T * c v dT 

The work is zero. 


1 _ c p dT v dP 
as y ~ JT 


c p dT 


RdP 
J P 


(18:3) 


' Ti c p dT 


T 1 


R [ P2 dP 
J p, P 


T2 c p dT R,__ P 2 


T 1 


jloge-pr (18:4) 










REAL GASES 


437 


3. Isothermal (n = 1). Table 9:2 may be used to find the final pres¬ 
sure, temperature, and volume. There is no change in internal energy 
since the internal energy is still a function of temperature alone. The 
heat flow and the work may also be based directly on the information 
presented in Table 9:2. 

4. Reversible adiabatic. This process follows a path defined as 
P\ k = const as before, but there is an important difference, for k is no 
longer a constant. This follows from the change which is now taking 
place in both c p and c v with temperature, while the difference between 
them remains constant. Table 9:2 therefore cannot be used to locate the 
final pressure, temperature, or volume, nor can the expression for work 
presented in that table be applied. The heat flow is zero by definition 
of the adiabatic, and this means, for the reversible process, that the 
entropy is constant. We may therefore write Eq. (18:3) in the form 


and, integrating, 


ds 


c p dT R dP 

T J P 



J [ T2 c p dT 

R J Tl T 


(18:5) 


This equation provides the means of establishing a relation between the 
change of pressure and the change of temperature. The perfect-gas rela¬ 
tion may now be used to fix the final volume. The work is equal to the 
decrease of internal energy and may be calculated by applying Eq. (18:2). 

5. Polytropic (n = const). As indicated in the heading, the polytropic 
for the perfect gas with variable specific heats is defined as a reversible 
process following a path represented by the equation PV n = const, with 
n a constant. No longer does this also mean that the amount of heat 
flow accompanying a unit change in temperature is the same throughout 
the process, as is the case when c p and c v are constant (see Art. 9:5). 
However, since the path is established, Table 9:2 may be used to find the 
final pressure, temperature, and volume, as for (1), (2), and (3) above. 
Also, again because the path is known, the use of that table for the calcu¬ 
lation of the amount of work is made possible. When Eq. (18:2) has been 
used to compute the change of internal energy, the heat flow is obtained 
from the simple addition of the work and the change of internal energy, as 
based on Eq. (2:3). 

The observant reader will have noted that the use of specific heats 
which are functions of the temperature has added little to the difficulty of 
solution of the reversible process for a perfect gas except in the evaluation 
[ T ' C T2 [ T - c dT 

of / CpdT, c v dT, and / -?-=— The amount of time and labor 

J Ti J T i J Ti 1 






438 


BASIC ENGINEERING THERMODYNAMICS 


required in the evaluation of these quantities depends upon the form of 
the expression for c p (c v , it will be remembered, differs only in the constant 
term) and explains why an expression of the form c v — A + BT is 
favored for many engineering calculations where extreme accuracy is not 
necessary. 

Example 18:2. A pound of air is adiabatically and reversibly compressed from an 
initial state at which p j = 14.7 psia, t\ = 40°F, to a final pressure p 2 of 75 psia. 
(a) Assuming the specific heat to be expressed by Eq. ( 8 ) of Table 18:1, calculate the 
final temperature, the change of enthalpy, internal energy, and entropy, and the work 
done on the air. ( 6 ) Compare with the same quantities as calculated by assuming 
constant specific heats as in Table 9:1. 

Solution: 

(a) From Eq. (18:5), 

, 75 , 778 [T* (6.37 + 0.000886T) dT 

l0g ”l4T7 = 163 “ 53^ jsoo -• 28.96 T - 

= (0.504) [6.37 log. ~ + 0.000836(7'; - 500) 

L 5UU 

Solving, 

T 2 = 797°R 
From Eq. (18:1), 

*• - h ' - 28796 C (6 ' 37 + 0 000S86D dT = (g^) [6.B7T + 0.0004432”® 

= ( 2 ^ 90 ) t 1892 + 170.5] = 71.2 Btu 
mc v = 4.385 + 0.000886 T 
From Eq. (18:2), 


1 7797 

u 2 - ui = / (4.385 + 0.0008867 7 ) dT 

28.96 J 500 

= t 1307 + 170 - 5 1 = 51.0 Btu 



[4.385 T 7 + 0.000443 7’ 2 ] 


From Eq. (18:4), 


S 2 Si 


1 f 797 (6.37 + 0.000886 T) dT 

28.96 J 500 T 

6.37 797 (0.000886) (297) 

28.96 ge 500 + 28.96 


53.3. 75 

778 ° §c 14.7 

(0.0685) (1.63) = 0.103 + 0.009 


1 W 2 

Cb) T 2 


= 0 (isentropic) 

J(ui - u 2 ) = (778)( — 51.0) = -39,700 ft-U 

. 0c—l)/k 


-'■(£)' 


= 500 


/ 75 \° 288 
V147,/ 


799°R 


- 0.112 


h 2 - hi = 0.24(799 - 500) = 71.7 Btu 
u 2 - m = (0.171) (299) = 51.2 Btu 

«2 — Si = 0 

iW 2 = (778)(—51.2) = -39,800 ft-lb 


















REAL GASES 


439 


The discrepancy between the two sets of results averages less than 1 per cent. How¬ 
ever, if the temperature range had extended into the higher temperatures the differ¬ 
ence would have been greater. 

18:3. The Gas Tables. Many engineering problems are concerned 
with the expansion or the compression of gases. The gas tables 1 make 
less tedious the solution of these problems by compiling the properties of 
air, of the products of combustion of a hydrocarbon of formula (CH 2 ) n 
with 400 and 200 per cent excess air, 2 of nitrogen, oxygen, water vapor, 
carbon dioxide, hydrogen, and carbon monoxide in a form designed to 
be of greatest advantage to the engineer in making the usual class of 
computations. 

The basic assumption in the compilation of the gas tables is that the 
gases are, at low pressures, 3 perfect gases with variable specific heat. The 
equation for the specific heat is based on spectroscopic data; it is a more 


Table 18:2. Air at Low Pressures* 


T 

t 

h 

Vr 

u 

V r 


300 

-159.7 

71.61 

0.17795 

51.04 

624.5 

0.46007 

400 

-59.7 

95.53 

0.4858 

68.11 

305.0 

0.52890 

500 

40.3 

119.48 

1.0590 

85.20 

174.90 

0.58233 

520 

60.3 

124.27 

1.2147 

88.62 

158.58 

0.59173 

540 

80.3 

129.06 

1.3860 

92.04 

144.32 

0.60078 

560 

100.3 

133.86 

1.5742 

95.47 

131.78 

0.60950 

600 

140.3 

143.47 

2.0050 

102.34 

110.88 

0.62607 

700 

240.3 

167.56 

3.446 

119.58 

75.25 

0.66321 

795 

335.3 

190.59 

5.404 

136.09 

54.49 

0.69405 

800 

340.3 

191.81 

5.526 

136.97 

53.63 

0.69558 

868 

408.3 

208.41 

7.389 

148.91 

43.51 

0.71550 

900 

440.3 

216.26 

8.411 

154.57 

39.64 

0.72438 

1000 

540.3 

240.98 

12.298 

172.43 

30.12 

0.75042 

1500 

1040.3 

369.17 

55.86 

266.34 

9.948 

0.85416 

2000 

1540.3 

504.71 

174.00 

367.61 

4.258 

0.93205 

3000 

2540.3 

790.68 

941.4 

585.04 

1.1803 

1.04779 

4000 

3540.3 

1088.26 

3280 

814.06 

0.4518 

1.13334 

5000 

4540.3 

1392.87 

8837 

1050.12 

0.20959 

1.20129 

6000 

5540.3 

1702.29 

20120 

1291.00 

0.11047 

1.25769 


* Abstracted by permission from Table 1, “Gas Tables’’ by J. H. Keenan and J. Kaye, John Wiley & 
Sons, Inc., New York, 1948.’ 


1 J. H. Keenan and J. Kaye, “Gas Tables,” John Wiley & Sons, Inc., New York, 
1948. 

2 See Art. 19:2. 

3 Low pressures are assumed to include the pressures normally encountered in the 
engineering process. The error of the air table, at 32 °F, is about 1 per cent at 300 psia, 
it is less at lower pressures or higher temperatures. For gases with higher critical 
temperatures than air, the error is greater than for air at the same pressure and 
temperature. 


















440 


BASIC ENGINEERING THERMODYNAMICS 


accurate and more complicated expression than the engineer would be apt 
to choose for his purpose if he were required to make the necessary detailed 
computations himself. The tables are keyed by absolute temperature in 
degrees Rankine and cover various ranges according to the gas; the range 
for the air table is 100 to 6500°R. The base temperature, at which the 
enthalpy is zero, is 0°R. Since the product PV is zero at this tempera¬ 
ture, the internal energy is also zero. The properties of air are given with 
respect to 1 lb, for the other gases in terms of the mole. 

An abstract from Table 1 of the gas tables, the air table, is presented 
here by permission as the basis of a discussion of the method of use of the 
tables, all of which are similarly arranged; it is Table 18:2. 

The enthalpy is tabulated in the third column of the air table. It has 
been computed by applying Eq. (18:1) with hi = 0 at 0°R, or 

h = fj c„dT (18:6) 


The fourth column tabulates values of the relative pressure p r . When 
the values of p r are compared at two temperatures, they furnish a key to 
the pressure ratio in an isentropic process between the limiting tempera¬ 
tures. Thus, denoting the two temperature levels by the subscripts 1 
and 2, 



(18:7) 


The fifth column of the air table lists values of the internal energy. 
Since u = h — Pv/J = h — RT/J, the internal energy may be readily 
calculated from the enthalpy but is here tabulated for convenience. 

Values of the relative volume v r are tabulated in the sixth column. As 
was the case with p r , the ratio of the values of v r at any two temperatures 
equals the ratio between the volumes at the ends of an isentropic that 
connects the two temperatures, or 



(18:7a) 


Any value may be arbitrarily assigned to either p r or v r at any one tem¬ 
perature, the values at all other temperatures being based upon the selec¬ 
tion which is decided upon. The values which are assigned in the air 
table are such that at each temperature v r is numerically equal to the vol¬ 
ume of 1 lb, in cubic feet per pound, at the pressure p r , in psia. The rela¬ 
tion between v r and p r is therefore v r = 53.3T/144p r . 

The last column of the air table gives values of 

. _ f T c p dT 


o 


(18:8) 




REAL GASES 


441 


This quantity, it will be noted, is also a property that is a function of the 
temperature alone; its value is zero at the base temperature of 0°R. 
From Eq. (18:14), 


s 2 Si — 



R 

J 


log. 


P2 

Pi 


= <f >2 ~ 0i — ylog e — (18:9) 

j p i 


The tables of properties of the other gases differ from the air table only 
in that h, u, and <j> are stated for 1 mole instead of 1 lb and that v r is the 
volume in cubic feet per mole at the pressure p r , so that v r = 1 5457yi44p r . 
This change is convenient in dealing with hydrocarbons and the products 
of their combustion since it makes it possible for a single table to serve for 
a whole series of hydrocarbons. 

The data of the gas tables may be applied to good effect in calculations 
concerned with any of the types of processes discussed in Art. 18:2 since 


CT 2 CTi 

Ah may be substituted for / c p dT, A u for / c v dT, and A</> for 
' Ti c v dT 

A great deal of additional information is also tabulated in 


Ti 


T 


the gas tables; some of this concerns other properties, such as the coeffi¬ 
cient of viscosity, and some is concerned with smoothing the path of the 
engineer in special calculations, such as the plotting of Fanno and 
Rayleigh lines. Two simple examples of the application of the air table 
to the solution of an engineering problem are given below. 


Example 18:3A. Solve Example 18:2, using the air table. 

Solution. At Ti = 500°R read p ri = 1.0590. Then p r2 = (75/14.7) (1.0590) = 
5.40. This corresponds, closely, to the tabulated value of p r at T 2 = 795°R. The 
change of enthalpy is the difference between the tabulated values of the enthalpy at 
795°R and 500°R, or 


h 2 - h x = 190.59 - 119.48 = 71.11 Btu 


Using the tabulated values of the internal energy, 
u 2 - Ux = 136.09 - 85.20 = 50.89 Btu 
Checking the change of entropy, which should be zero for this isentropic process, 


Sz _ Sl = H - <*,! - | log, 21 = 0.69405 - 0.58233 - (^||) (1.63) = 0 
Also, 

X W 2 = (778)(-50.89) = -39,600 ft-lb 

Checking the ratio of the tabulated values of v r , v ri = 174.90, and v n = 54.59. 
Therefore 


V 2 _ £r2 
V l Vri 


54.49 

174.90 


0.312 






442 


BASIC ENGINEERING THERMODYNAMICS 


From Pv = RT, 

(53.3) (795) 
" 2 (75) (144) 

(53.3) (500) 
Vl (14.7) (144) 

and 


3.93 ft 3 /lb 
12.6 ft 3 /lb 


V 2 

Vi 


3.93 

12.6 


0.312 


Example 18:35. Air is compressed adiabatically from 14.7 psia, 40°F, to p 2 — 75 
psia in an axial-flow compressor with an efficiency (see Art. 11:15) of 80 per cent. 
Based on the air table, find the final temperature and the work of compression per 
pound of air compressed. 

Solution. The actual change of enthalpy is, according to the definition of com¬ 
pressor efficiency given in Art. 11:15, equal to the isentropic change of enthalpy 
between the initial state and the final pressure, divided by the efficiency. The isen¬ 
tropic change of enthalpy has been shown in Example 18:3A to be 71.11 Btu. Then 

h 2 - hi = = 88.9 Btu and h 2 = hi + 88.9 = 208.38 Btu 

U.oU 

This enthalpy corresponds, according to the air table, to a temperature T 2 = 868°R. 
The work of compression for this adiabatic steady-flow process is 

x Wi = J{h x - h 2 ) = (778) (-88.9) = -69,200 ft-lb/lb 


18:4. The van der Waals Equation. Based on the equation of state of 

a perfect gas, the pressure 


P = 


RT 

v 


The concept of a perfect gas (see Art. 9:2) is of a hypothetical substance 
the molecules of which are so far apart that intermolecular collisions do 
not occur and attractive forces between molecules vanish; consistent with 
these two assumptions, we find that the perfect-gas relation holds with 
greater accuracy at lower pressures and, consequently, larger volumes. 

The effect of intermolecular collision is to cause a given molecule to 
strike the walls of the container more frequently and thus, when the total 
effect of the bombardment of all the molecules is considered, to increase 
the pressure on those walls. In 1873, van der Waals advanced the theory 
that the pressure would, owing to this effect, be increased in inverse pro¬ 
portion to the ratio of the net volume, obtained by subtracting the actual 
volume of all the individual molecules from the total space occupied by 
the gas, to the total volume which the gas fills. As the gas nears the 
dense liquid phase, this ratio approaches a minimum and the increase of 
pressure due to this effect a corresponding maximum. 

The effect of the attractive forces between molecules is, in the limit as 
the liquid phase is approached, to cause the molecules to cling together; 







REAL GASES 


443 


the outward pressure exerted against the walls of a container is reduced. 
Van der Waals suggested that the pressure of the gas would be reduced, 
because of this effect, in proportion to the square of its density and pro¬ 
posed the following equation to reflect both effects in altering the pressure, 


RT _ a 
v — b v 2 


(18:10) 


in which a and b are dimensional constants. The constant 6 represents 
the volume actually occupied by the molecules of the substance, and its 
size is therefore of the order of the specific volume of the fluid in its liquid 
phase, with the molecules closely packed together; v — b is correspond¬ 
ingly the net volume to which refer¬ 
ence has been made above. At low 
pressures, v is large, and v — b ap¬ 
proaches v in magnitude; as a result 
the RT/(v — b) term approaches in 
value the pressure as determined 
from the perfect-gas relation. The 
constant a is a proportionality con¬ 
stant which reflects the effect of 
density in reducing the pressure by 
accounting for an increase in the at¬ 
tractive forces between molecules. 

At low pressures, v 2 becomes very 
large and the reduction of pressure 
represented by the a/v 2 term corre¬ 
spondingly small. Thus the van der 
Waals equation approaches the per¬ 
fect-gas relation at low pressures. 

The van der Waals equation may be easily solved for pressure when the 
values of the constants and of the temperature and volume are known. 
In the form 



Fig. 18:1. The van der Waals equation 
of state. 


RT 


= (p+£)(»-V 


(18:11) 


it may be as readily solved for temperature; for volume alone it is not 
explicit. With known values of a and b, Eq. (18:10) may be used to plot 
the family of isothermals shown in Fig. 18:1. Below a certain tempera¬ 
ture Tc, these isothermals will be observed to have three real roots in v in a 
fixed range of pressures (for the isothermal T = T i, within the range Ph 
to Pj). Above T c there is only a single real root for every value of P, 
and the curve (T = T $) has a negative slope throughout while, at 1 c , the 
isothermal, although it has everywhere only a single root in v, is instan- 








444 


BASIC ENGINEERING THERMODYNAMICS 


taneously horizontal at c. If the points e, d, c, f, and g are now connected, 
as by the dashed line of the figure, the resemblance to the line aimgnjc of 
Fig. 7:3 is noteworthy. On Fig. 7:3 this line connects saturated-liquid 
and saturated-vapor states on a pressure-volume diagram. 

The resemblance of isothermal T 3 of Fig. 18:1 to isothermal rs of Fig. 
7:3 is marked, as is that of isothermal T c of Fig. 18:1 to the isothermal at 
the critical temperature in Fig. 7:3. Below T c the isothermals of the two 
figures are also similar in appearance except in the two-phase region 
between the saturated-liquid and saturated-vapor lines. In this space, 
the isothermal of the substance that is in stable equilibrium becomes 
horizontal while the van der Waals isothermal is a rising and falling curve. 
The dotted line eig has been added on Fig. 18:1 to represent the state path 
followed by the two-phase fluid when the temperature is constant; the 
constant pressure along this path is the saturation pressure equivalent to 
the temperature TV 

With reference to Fig. 18:1, points along the path segment eh represent 
states at which the pressure of the liquid is less than the saturation pres¬ 
sure equivalent to its temperature. These are metastable states that 
may be observed experimentally under laboratory conditions; the liquid 
is said to be superheated. Similarly, the path segment gj passes through 
states at which the temperature of the vapor is below the saturation tem¬ 
perature corresponding to its pressure; this class of metastable states is 
often observed in the flow of steam through nozzles and has been discussed 
in Art. 11:8. On the other hand, states along the path segment hij, 


where 



> 0, may be shown, by methods which are beyond the 


scope of this text, to be unstable; they will not exist long enough to be 
observed. 

The purpose of the preceding paragraph has been to indicate that the 
undulating curve of the van der Waals isothermal as it crosses the two- 
phase region, although it does not connect states of stable equilibrium, 
may be interpreted in part in terms of observable states of the real fluid. 
In locating the stable-equilibrium line eig across the two-phase region, it 
is observed that this line is a line of constant pressure and of constant 
temperature. A cycle may be reversibly traversed by following the path 
ehijgie, and, during this cycle, the temperature has remained constant. 
According to the Second Law, no net work could have resulted, and area 
eih must therefore be equal to area ijg. 

An obvious point of coincidence between Figs. 7:3 and 18:1 is the point 
c at which the isothermal T c of Fig. 18:1 becomes instantaneously horizon¬ 
tal; this corresponds to the critical point g of Fig. 7:3. For many real 
fluids the critical pressure, temperature, and volume have been deter¬ 
mined, even though their equations of state have not been derived, and 



REAL GASES 


445 


this ciitical state can be made the basis of the selection of the values of the 
constants a and b of the van der Waals equation when that equation is to 
be fitted to a particular substance. 

If the first derivative of the van der Waals equation is placed equal to 
zero, the result will be the equation of the line which passes through the 
minima and maxima of the family of isothermals pictured in Fig. 18:1 
(connecting points such as h, k, c, l, and j). Thus 



-RT 
{v — b ) 2 


+ ^ = 0 
v 6 


(18:12) 


is the equation of this line. The maximum on the curve represented by 
this equation occurs at critical point c and may be located by taking the 
first derivative of Eq. (18:12), which is the second derivative of the van 
der Waals equation, and equating it to zero, or 


a 2 /A = 2 RT _ 6 a 
dv 2 ) T (v — b) s v A 


(18:13) 


But point c satisfies both Eq. (18:12) and Eq. (18:13) and, solving them 
simultaneously, 


Substituting in Eq. (18:10), 


v c = 35 

rn 

c ~ 27bR 

(18:14) 

(18:15) 

p _ ® 

c 27 b 2 

(18:16) 


Table 18:3. Critical Constants* 


No. 

Gas 

Sym¬ 

bol 

Mol. wt 

1 

Oxygen 

o 2 

32.00 

2 

Nitrogen 

n 2 

28.02 

3 

Air 

• • • • 

28.96 

4 

Hydrogen 

h 2 

2.016 

5 

Carbon dioxide 

co 2 

44.00 

6 

Carbon monoxide 

CO 

28.00 

7 

Water 

h 2 o 

18.02 

8 

Methane 

ch 4 

16.03 

9 

Ethylene 

c 2 h 4 

28.03 

10 

Ethane 

c 2 h 6 

30.05 

11 

Octane 

CgHis 

114.1 


R, 

ft/°R 

To, 

°R 

Vc, 

psia 

V c , 

ftVlb 

Z c 

PcVc/RT c 

48.3 

278 

730 

0.0372 

0.292 

55.2 

227 

492 

0.0515 

0.292 

53.3 

238 

547 

0.0470 

0.292 

766 

60 

188 

0.517 

0.305 

35.1 

548 

1073 

0.0348 

0.279 

55.2 

241 

508 

0.0515 

0.283 

85.8 

1165 

3206 

0.0503 

0.232 

96.4 

344 

673 

0.0989 

0.289 

55.1 

509 

747 

0.0728 

0.279 

51.4 

550 

708 

0.0763 

0.275 

13.5 

1025 

362 

0.069 

0.260 


* Data from various sources, including International Critical Tables, and Natl. Bur. Standards (U.S .) 
Circ. 279 (December, 1925). 

































446 


BASIC ENGINEERING THERMODYNAMICS 


The ratio Pv/RT is called the compressibility factor and is denoted as Z. 
For the perfect gas, it has the constant value of 1; for real gases it is not a 
constant but is a function of both temperature and pressure. At the 
critical point the value of the compressibility factor is 

Zc = ~ c (18:17) 

When the values of the critical volume, temperature, and pressure from 
Eqs. (18:14) to (18:16) are substituted in Eq. (18:17), it is discovered that 
the van der Waals equation gives a value of Z c = f. In Table 18:3 are 
presented critical-state data for a number of substances. It is immedi¬ 
ately observed that Z c for all these substances is well below the value of f 
which is obtained for that ratio from the van der Waals equation, and it 
appears that the application of the equation at states close to the critical 
point will result in errors of the order of 10 to 40 per cent. 1 

In spite of its lack of accuracy, the van der Waals equation is of con¬ 
siderable interest and importance because: 

1. It illustrates a rational method of attacking the problem of setting 
up a general equation of state. 

2. Its lines of constant temperature, even in the two-phase region where 
they seem to be widely different from the isothermals of a vapor, may be 
interpreted in terms of the behavior of the real substance. 

3. It covers all fluid regions, from the compressed liquid to the gas, and 
shows the general characteristics of the real substance, even though it does 
not give precise values of the properties. 

No single general equation of state is known which represents the proper¬ 
ties of all fluid phases with precision. Assuming that such an equation is 
developed, it is highly probable that it will be of such complexity as to be 
of little practical use to the engineer. 

The critical pressure and temperature can be more easily and accurately 
measured than can the critical volume. For this reason, although Eq. 
(18:14) can be readily solved for b in terms of v c , it is better to express its 
value in terms of T c and P c ; the same is true of* the constant a. Thus, 
from Eq. (18:16), 

a = 27 b 2 P c (18:18) 

1 The van der Waals equation can be made to give a better fit for states near the 
critical state by assigning an artificial value to R such that Eq. (18-17) will yield the 
compressibility factor that is characteristic of the particular substance at its critical 
state. However, when this is done, the equation loses accuracy when applied to 
other areas. For example, at low pressures it no longer approximates the perfect-gas 
relation, which is known to hold closely in that region. 



REAL GASES 


447 


and, substituting this value in Eq. (18:15), 


or 


T c = 


8bP c 

R 



Substituting this value of b in Eq. 

a = 


(18:18), 

27 R 2 T C 2 
64P C 


(18:19) 


(18:20) 


Example 18:4. Using the van der Waals equation, calculate, for water, the pressure 
corresponding to temperature-volume conditions a, b , and c, as given below. At the 
calculated pressure and the given temperature, read the volume from the steam 
tables and compare with the given volume, finding the error percentage, (a) t = 
1200°F; v = 0.24 ft 3 /lb. (6) t = 720°F; v = 0.17 ft 3 /lb. (c) t = 500°F; v = 40ft 3 /lb. 

Solution: 


(a) From Eq. (18:19), 

_ (85.8) (1165) 

(8) (3206) (144) 

From Eq. (18:20), 

(27)(85.8) 2 (1165 2 ) 
a (64) (3206) (144) 


= 0.0271 ft 3 /lb 


= 9130 


(psf)(ft 6 ) 

lb 2 


Substituting T a and v a in Eq. (18:10), 


(85.8) (1660) 
0.24 - 0.0271 


9130 

0.24 2 


= 669,000 - 158,500 = 510,500 psf = 3540 psia 


At this pressure and the temperature of 1200°F, the steam tables give the specific 
volume as 0.2514. At this state the van der Waals equation gives too low a volume 
by about 5 per cent. 


(85.8) (1180) 
0.17 - 0.0271 


9130 

0.17 2 


= 708,000- 316,000 = 392,000 psf = 2720 psia 


The steam-table value for the volume at this pressure and a temperature of 720°F is 
0.1569. The van der Waals equation gives a volume that is too high by about 
9 per cent. 


(85.8) (960) 
40 - 0.0271 


9130 

40 2 


= 2040 — 6 = 2034 psf = 14.1 psia 


At p = 14.1 psia, t = 500°F, the steam tables give a specific volume of 40.44 ft 3 . 
The van der Waals volume is about 1 per cent low. 

States a, b, and c in this example have been chosen to illustrate the relative accuracy 
to be expected from the use of the van der Waals equation in, respectively, the high- 
pressure high-temperature range, the range of states close to the critical point, and 














448 


BASIC ENGINEERING THERMODYNAMICS 


the low-pressure area. The calculations above indicate, as was to be expected, that 
the region to which it applies with the least accuracy is in the neighborhood of the 
critical state. Exactly at the critical state the van der Waals equation gives a volume 
which is 0.375/0.232 — 1 = 0.62, or 62 per cent too high. 


18:5. Other General Equations of State. A number of general equa¬ 
tions of state have been proposed, some of which follow the van der Waals 
equation in basing their development on the kinetic theory of gases while 
others are empirically derived to agree with observed data. Only a few 
will be discussed here. 

1. The Dieterici equation was first proposed in 1899. It gives a value 
of Z c = 0.27, a better average value for real gases than the 0.375 of the 
van der Waals equation. According to this equation in its revised form 

P = ^N-e ¥Ur " ( 18 : 21 ) 

v — b 

with a = 4:RT C 2 - 27 /P c e 2 and b = RT c /P c e 2 ; e is the natural logarithmic 
base. 


Example 18:5A. Apply the Dieterici equation to water at the states suggested in 
Example 18:4, and determine the deviation from steam-table volumes as in that 
example. 

Solution: 


b 


a 


(85.8) (1165) 
(3206) (144) (2.718) 2 
(4)(85.8)(1165) 2 - 27 
(3206) (144) (2.718) 2 


0.0294 

916 


(a) Pa 


(85.8) (1660) (1660 1 - 27 ) (0.24) 

0.24 - 0.0294 


676,000e~°- 313 = 495,000 psf = 3440 psia 


At p = 3440, t = 1200, the steam tables give a volume of 0.2596. The Dieterici 
volume is about 7 per cent too low. 


(b) P b 


(85.8) (1180) (ii8o i27 )(o.i7) 

0.17 - 0.0294 


720,000e 0 677 = 366,000 psf = 2540 psia 


At p 2540, t /20, the steam-table volume is 0.179. The Dieterici volume is 
in error by about 5 per cent. 


(c) Pc 


(85.8) (960) (iisoi 27)(40) 
40 - 0.0294 


2062e-° 0037 = 2055 psf = 14.27 psia 


The steam tables give v = 39.98 at p = 14.27, t = 500. This agrees almost exactly 
with the Dieterici volume. 


2. The Berthelot equation, proposed in 1903, resembles the van der Waals 
equation and gives the same value of Z r as that equation. It is suitable 













REAL GASES 


449 


for use at states having temperatures well above the critical temperature. 


RT _ _a L 
v — b Tv 2 


(18:22) 


with a = 27P 2 T C 3 /64P C and b = RT C /SP C . 

3. The Beattie-Bridgman equation (1928) is an example of an equation 
that has been devised to fit observed experimental data. Its constants 
cannot be evaluated from critical-state data alone but must be taken from 
tables which give their value for specific gases. It therefore cannot be 
classified with the van der Waals, Dieterici, and Berthelot equations as 
an entirely general equation of state. On the other hand, as would be 
expected, it is the most accurate of the four in the range in which its use is 
recommended; it should not be used when the specific volume is less than 
twice the critical specific volume. The equation is stated as 


P = 


mRT( 1 — e) 
m 2 v 2 


(mv + B) 


A 

m 2 v 2 


(18:23) 


Table 18:4. Constants for Beattie-Bridgman Equation of State* 


No 

Gas 

Symbol 

Ao , 

(psf)(ft 6 ) 

mole 2 

a, 

ft 3 

mole 

Bo , 

ft 3 

mole 

b, 

ft 3 

mole 

c X 10- 6 
(ft 3 )(°R) 3 
mole 

1 

Oxygen 

o 2 

808,000 

0.410 

0.740 

0.067 

4.48 

2 

Nitrogen 

n 2 

730,000 

0.419 

0.808 

-0.111 

3.92 

3 

Hydrogen 

h 2 

107,000 

-0.081 

0.336 

-0.698 

0.0471 

4 

Carbon dioxide 

co 2 

2,720,000 

1.143 

1.678 

1.159 

61.65 

5 

Carbon monox¬ 
ide 

CO 

729,000 

0.419 

0.808 

-0.111 

3.92 

6 

Methane 

ch 4 

1,236,000 

0.297 

0.894 

-0.254 

11.98 

7 

Ethylene 

c 2 h 4 

3,330,000 

0.794 

1.945 

0.575 

21.18 

8 

Ethane 

c 2 h 6 

3,190,000 

0.938 

1.504 

0.306 

84.20 


* From Proc. Am. Acad. Arts Sci., 63 , 64 (1928—1930). 


where A = A 0 ( 1- —\ B = B 0 ( 1- \ e — — 7 ™? an d Ao, a, B 0 , b, 

\ mv) \ mv / mv 1 

• 

and c are constants having different values for each gas. Table 18:4 gives 
their values for a few substances. In this equation m is the molecular 
weight, mv being the molar volume and mR the universal gas constant. 

Example 18:511. Using the Beattie-Bridgman equation, calculate the pressure 
that corresponds, for carbon dioxide, to a temperature of 250 F and a specific volume 
of 0.2 ft 3 /lb. Using the calculated pressure, determine the value of Z , the com¬ 
pressibility factor. 
































450 


BASIC ENGINEERING THERMODYNAMICS 


Solution. From Eq. (18:23), using the constants for C0 2 from Table 18:4, 


A = 2,720,000 [l - ( 4 ff ( f 2 ) ] = 2,350,000 

B = L678 [ 1 “( TO )] =L457 


61,650,000 

e ~ (44) (0.2) (710) 3 “ 
mv = (44) (0.2) = 8.8 


0.0196 


(1545) (710) (1 - 0.0196) (8.8 + 1.457) - 2,350,000 
^ ~ 8.8 2 


= 112,500 psf = 780 psia 


Pv _ (780) (144) (0.2) 
RT (35.1)(710) 


0.90 


4. For none of the foregoing equations, including the equation of van 
der Waals, is the volume an explicit function. Since the pressure and the 
temperature are the properties most readily and most accurately measur¬ 
able, an expression for which v is the dependent variable is desirable. 
Such an equation is 

v = ^ (1 + aiP + 2a<P‘) (18:24) 


in which ZaJP 1 represents a series of terms in each of which a is, as is a h 
a function of temperature alone and the exponent i is in each term greater 
than 1. It is usual, though not necessary, to use integer powers of P. A 
formula of this type has been used in the preparation of the Keenan and 
Keyes steam tables. 1 It will be observed that at zero pressure a value of 
Z = 1 will result, as for the perfect gas and all of the general equations of 
state discussed above. 

By mathematical manipulation, many equations of state for which P is 
the dependent variable may be closely approximated in the often more 
convenient form in which v is explicit; a series results which follows Eq. 
(18:24) in its form. 2 

18:6. The law of corresponding states was proposed to meet the situa¬ 
tion where detailed information, with the exception of the values of the 
pressure, temperature, and volume at the critical state, in regard to a 
particular substance is lacking but a fairly close approximation of its 
properties is acceptable. The law is based on the premise that the PvT 
surfaces (see Fig. 7:1) of all substances are geometrically similar, and may 
be stated in mathematical terms as 


F* = }{P r ,Tr) 


(18:25) 


1 See J. H. Keenan and F. G. Keyes, “Thermodynamic Properties of Steam/’ 
equation (13), p. 15, John Wiley & Sons, Inc., New York, 1936. 

2 For an application of this principle to the Beattie-Bridgman equation, see J. H. 
Keenan, “Thermodynamics,” John Wiley & Sons, Inc., 1941. 









REAL GASES 


451 


in which V R , P R , and T R are the reduced values of the corresponding prop¬ 
erties, i.e. f the ratio of their values to the value of the corresponding prop¬ 
erty at the critical state. Thus V R = v/v c , P R = P/P e , and T R — T/T c . 
If two fluids have the same value of V R , P R , and T R , they are in corre¬ 
sponding states. 

In making use of this law, when it is desired to determine the volume at 
some state where the pressure and temperature are, respectively, P and T, 
the values of the ratios P R and T R are computed. These ratios are then 
carried to a chart or table which shows the P, T , and v relations for some 
second substance, whose properties are known with a high degree of pre¬ 
cision, and the corresponding state is located. At that state the volume 
is obtained and compared with v c for this reference substance. This gives 
V R , and this ratio may now be carried back and used in connection with 
v c for the first substance to obtain its volume at the stated pressure and 
temperature. 

The law of corresponding states is not suitable for use at low pressures 
where all gases obey the perfect-gas relation. For the law assumes that 
the values of Z, the compressi¬ 
bility factor, for the two gases at 
corresponding states will always 
be in the ratio of the values of 
this factor at their respective 
critical states. At zero pressure 
Z = 1 for all gases, and, for the 
law to hold, it would be required 
that the ratio Z C JZ C „ where the 
subscripts 1 and 2 refer to the two 
gases that are being compared, 
must be 1. In Table 18:3 it is 
shown that all gases do not have 
the same ratio Z c , the variation 
being from about 0.2 to 0.3. The 
law gives fairly accurate results 
when used for states near the 
critical state; it is also used in 
the range where T R and P R are 
greater than 1. Figure 18:2 pre¬ 
sents a chart which may be used for this purpose; it shows the reduced 
isometrics on a graph for which P R is the ordinate and T R the abscissa. 
The isometrics shown on this chart are average curves for a number of 
gases, all of which have values of Z c in the vicinity of 0.28. 

18:7. The Generalized Z Chart. The relation between Z and the 
reduced pressure is similar for different substances, and this similarity 



Fig. 18:2. The law of corresponding states. 
(From thesis by W. C. Kay, Massachusetts 
Institute of Technology, 1937.) 



















452 


BASIC ENGINEERING THERMODYNAMICS 



0 1.0 2.0 3.0 4.0 5.0 6.0 70 


Reduced pressure, P R 

Fig. 18:3. Generalized compressibility factor—reduced pressure chart. [Adapted 
from Gonq-Jen Su, “Modified Law of Corresponding States for Real Gases,” Ind. Eng. 
Chem., 38 (August, 1946).] 



Fig. 18:4. Z chart for high ranges. {Adapted from chart by Weber from data of Gonq- 
Jen Su, “ Thermodynamics for Chemical Engineers,” by Harold C. Weber, John Wiley & 
Sons, Inc., New York, 1939.) 





































REAL GASES 


453 


may be used as the basis of an estimate of the volume of a gas when infor¬ 
mation as to its critical state is available. Generalized charts on which 
the values of Z are plotted as ordinates against P R as abscissas are shown 
in Figs. 18:3 and 18:4. In using these charts the values of P R and T R that 
correspond with a given state at which it is desired to estimate the volume 
are calculated, the corresponding point is located on the chart, and the 
value of Z is read. This value of Z may then be used to calculate the 
volume. It will be observed that this method, although based in general 
on the law of corresponding states, avoids some of the inaccuracy of that 
law at low pressure, since all of the reduced isothermals pass through 
Z = 1 at zero pressure. 

Example 18:7. Using the generalized Z charts, check the value of Z that was 
calculated in Example 18:5 B. 

Solution: 

Pr = twts = 0.70; T r = HI = 1.295 

This point, when located on Fig. 18:3, gives Z = 0.89. This compares with Z = 0.90, 
as based on the Beattie-Bridgman equation. 

18:8. The Properties of a Real-gas Mixture. In any mixture of real 
gases that is in equilibrium, we may write 

7 7 m s'' rn rji r/i 

1 — 1 2 — i 3 — ••• — 1 n — 'm 

7 , = f 2 = F 3 = • • • = V n = V m 

Pi P 2 P 3 ~h " " ' T Pn — Pm 

in which the subscripts 1 to n refer to the components and the subscript m 
to the mixture. But the mole fractions of the constituents are not in pro¬ 
portion to their partial pressures in the mixture (see Art. 10:3) unless the 
pressure of the mixture is low and the mixture can be considered to be 
composed of perfect gases. At high pressures considerable error is 
invited in attempting to apply these mixture relations to the problem of 
approximating the connection between the pressure, volume, and tem¬ 
perature of a real-gas mixture. A number of other methods which seem 
more promising have been suggested, two of which are briefly discussed 
below. 

1 . Kay 1 assumes that the mixture can be treated as a single gas with a 
critical pressure and a critical temperature that have been obtained by 
averaging these values for the constituents of the mixture. The psuedo- 
critical temperature and pressure are based on the following equations, 


T Cm = Xi T Cl + x 2 T Ci + xzT Ci + 
P Cm = Xi P Cl + x 2 P Cl + XzP C3 + 


( 18 : 26 ) 


1 Ind. Eng. Chem., 28, 1014 (1936). 


454 


BASIC ENGINEERING THERMODYNAMICS 


in which the subscripts 1 , 2 , 3 , etc., refer to the constituents, the subscript 
m to the mixture, and x h x 2 , etc., are the respective mole fractions. When 
the pseudocritical temperature and pressure have been computed, they 
may be used to find the values of the constants in general equations of 
state, such as those of van der Waals, Dieterici, or Berthelot, or the 
generalized Z chart may be used if only a limited number of states are to 
be investigated. 

An illustration of this method of finding the critical pressure and tem¬ 
perature of a mixture is found in the data for oxygen, nitrogen, and air in 
Table 18:3. Air is essentially a mixture of oxygen and nitrogen with 
respective mole fractions equal to about 0.21 and 0.79. When Eqs. 
(18:26) are applied, the pseudocritical temperature and pressure are cal¬ 
culated as 238°R and 542 psia, checking the tabulated values for air 
closely. 

2. Beattie 1 has proposed that the Beattie-Bridgman equation of state 
may be used with the constants weighted as follows: 

A 0m = (aq A 0 * + x 2 A 0s * + x 3 Ao 3 i + • • -) 2 
a m = Xidi + x 2 a 2 + x 3 a 3 + • • • 

B 0m = X\Bq 1 + x 2 Bq 2 + x 3 B o 3 + • • • (18:27) 

b m = xibi + x 2 b 2 + x 3 b 3 + • • • 

C m X\C\ ~|- X 2 C 2 —(— X 3 C 3 ~ J— 

This method may be used when the Beattie-Bridgman constants are 
known for all of the constituents of the mixture and the mass per unit 
volume of each constituent is less than one-half the critical density of that 
constituent. 

18:9. Conclusion. In the foregoing articles of this chapter, methods 
of developing a PvT relation for the real gas have been discussed. When 
this relation has been established, the methods of Art. 8:3 will apply in 
writing secondary equations of state to express the enthalpy, the internal 
energy, and the entropy. The van der Waals, Dieterici, Berthelot, and 
Beattie-Bridgman equations all express the pressure as a function of the 
temperature and volume. Equations (8:30) to (8:33) illustrate the man¬ 
ner in which these PvT relations can be used to write the secondary equa¬ 
tions for internal energy and for entropy. With the pressure, volume, 
and internal energy known, Eq. (2:8) may be applied to calculate the 
enthalpy. 

The calculation of internal energy, enthalpy, and entropy is much more 
convenient when the volume is explicit in the primary equation of state, 
as in Eq. (18:24). The secondary equations of state then follow Eqs. 
(8:29) and (8:35) in form. 


1 /. Am. Chem. Soc., 51, 1929. 


REAL GASES 


455 


Problems 

1. (a) Plot the specific heat at constant pressure per pound of air vs. the temper¬ 

ature in Fahrenheit degrees in the range 2G0 to 1440°F as based on Eq. (8) of Table 
18:1. (6) Plot c v for air on the same graph and over the same range, (c) Show c p 

for air on the same graph as based on Eq. (9). 

2. Over the temperature ranges that are specified in each case, plot c p for hydrogen 
as based on Eqs. (10) to (13) of Table 18:1 (three curves). Assume a range of 0 to 
2000°F for Eq. (11). Compare the three values of c v at 1000°F. 

3. Table 18:1 is based on the assumption that the perfect-gas relation Pv = RT 
may be applied to the gases listed and is therefore limited in its application to com¬ 
paratively low pressures. Why are no monatomic gases included in the listing? 

4. (a) Use Eq. (8) of Table 18:1 in calculating the change of enthalpy per pound 

of air as the temperature changes from 340 to 1040°F. ( b ) Calculate the change of 

internal energy over the same temperature interval. ( c ) What is the mean value 
of c p ? 

5. Based on Eq. (8) of Table 18:1, calculate the change of entropy per pound of 
air as its temperature changes from 340 to 1040°F and its pressure from 20 to 35 psia. 

6. During a reversible process that takes place at constant atmospheric pressure, 

the temperature of a pound of air decreases from 1040 to %40°F. (a) Based on 

Eq. (8) of Table 18:1, calculate the heat flow, the external work, and the changes of 
internal energy and entropy that accompany the process. (6) Repeat as based on 
Eq. (9) of Table 18:1. (c) Compare the answers to parts a and b with those which 

result when c p is assumed to be constant at 0.24. 

7. During a reversible constant-volume process that begins at atmospheric pressure 
and a temperature of 340°F, the temperature of 1 lb of air increases to 1040°F. Based 
on Eq. (8) of Table 18:1, calculate the heat flow, the work, and the changes of internal 
energy and entropy that accompany the process. Compare with the similar quanti¬ 
ties when a constant specific heat (see Table 9:1) is used. 

8. At a constant temperature of 340°F, the pressure of an air system decreases 
from 80 to 20 psia during a reversible process. Equation (8) of Table 18:1 will be 
assumed to give the variation of specific heat with temperature. Calculate the heat 
flow, the work, and the changes of internal energy and entropy that accompany the 
process. Compare with the result when the specific heat is a constant. 

9. The reversible adiabatic expansion of a pound of air begins at a pressure of 

200 psia and a temperature of 1040°F and ends when the temperature becomes 340°F. 
(a) Assuming that Eq. (8) of Table 18:1 gives the relation between c p and T, what 
is the final pressure? What are the initial and final volumes? (6) Calculate the 
heat flow, the work, and the changes of internal energy and of entropy that accom¬ 
pany the process, (c) What is the mean value of k as based on the mean values of 
c p and c v and as calculated from Eq. (9:28)? ( d ) Assuming constant specific heats 

as given in Table 9:1, calculate the heat flow, the work, and the changes of internal 
energy and of entropy, and compare with the answers to part b above. 

10. A pound of air is compressed reversibly with n = 1.3 from atmospheric pres¬ 
sure and 340°F until it reaches a temperature of 1040°F. Equation (8) of Table 18:1 
will be assumed to represent the relation between c p and T. ( a ) What are the initial 
and final volumes and the final pressure? (6) Calculate the heat flow, the work, 
and the changes of internal energy and of entropy that accompany the process, 
(c) What is the average value of c n , and what are its values at the beginning and at 
the end of the process? ( d ) Assuming constant specific heats as given in Table 9:1, 
calculate the volumes, the final pressure, the heat flow, the work, and the changes of 
internal energy and of entropy, and compare with the answers to parts a and b. 


456 


BASIC ENGINEERING THERMODYNAMICS 


11. Solve Prob. 6, using the air table. Compare with the answers to Prob. 6, 
and explain any discrepancies. 

12. Solve Prob. 7, using the air table. 

13. Solve Prob. 8, using the air table. 

14. Solve Prob. 9, using the air table. 

15. Solve Prob. 10, using the air table. 

16. In Art. 18:3 it is stated that the gas tables for hydrocarbons are compiled on 
the basis of 1 mole instead of 1 lb since one table can then serve for a whole series of 
hydrocarbons. How is this possible? 

17. For ethylene (C 2 H 4 ) calculate the values of the constants a and b of the van der 
Waals equation. 

18. Based on the van der Waals equation, calculate the data for, and plot on a 
pv diagram, the isothermals for ethylene at 240, 49, —160, and —260°F. 

19. Show the derivation of Eqs. (18:14) to (18:16) in detail. 

20. Show that the compressibility factor at the critical state is f- according to the 
van der Waals equation. 

21. At atmospheric pressure, 1 lb of ethylene is found to have a volume of 5 ft 3 . 
According to the van der Waals equation, what is its temperature? What is the 
compressibility factor? 

22. At a temperature of 140°F, the specific volume of ethylene is 0.23 ft 3 /lb. 
According to the van der Waals equation, what is its pressure? What is the com¬ 
pressibility factor? Compare with the compressibility factor in Prob. 21, and discuss. 

23. Find the volume and pressure coordinates of the maximum and the minimum 
on the lowest isothermal of Prob. 18. 

24. In Prob. 22, what is the pressure as based on (a) the Dieterici equation; (b) the 
Berthelot equation; ( c ) the Beattie-Bridgman equation? In each case, what value 
of Z applies to the state? 

25. According to the Keenan and Keyes tables, the specific volume of steam at a 
pressure of 5000 psia and a temperature of 1400°F is 0.2027 ft 3 /lb. What are its 
reduced pressure, reduced temperature, and reduced volume? What is the corre¬ 
sponding state (give pressure, temperature, and specific volume) for ethylene? 

26. Using Fig. 18:2 and the data given in Table 18:3, estimate the specific volume of 
ethylene at 1500 psia and 350°F. 

27. Use the generalized Z charts and the data given in Table 18:3 to estimate the 
specific volume of ethylene at 1500 psia and 350°F. Compare with the answer to 
Prob. 26. 

28. Use the generalized Z charts and the data given in Table 18:3 to estimate the 
specific volume of hydrogen at a pressure of 5000 psia and a temperature of 80°F. 

29. A mixture is composed of 0.4 mole fraction of ethylene and 0.6 mole fraction of 
ethane, (a) Estimate the volume that corresponds to a pressure of 1500 psia and a 
temperature of 350°F using Kay’s method and the generalized Z charts. ( b ) Using 

• the Beattie-Bridgman relation, calculate the pressure corresponding to a temperature 
of 350°F and a specific volume of 0.167 ft 3 /lb. 


a , b, c 


Cp 

Cy 


A, B , C, D, E 


e 

h 

J 


Symbols 

constants 

specific heat at constant pressure 
specific heat at constant volume 
constants 

natural logarithmic base 
specific enthalpy 
proportionality factor 


REAL GASES 


457 


k 

VI 

n 

V 

Vr 

P 

Pr 

Q 

R 

s 

s 

t 

T 

T r 

u 

V 
V r 

V 

Vr 

w 

x 

Z 

z c 

Greek Letters 


0 

Subscripts 

c 

m 

V 

s 

T 

v 


ratio of the specific heats, c p /c v 

molecular weight; weight of 1 mole 

a constant; the polytropic exponent 

pressure, psi 

relative pressure 

pressure, psf; pressure in general 

reduced pressure 

heat flow 

gas constant 

specific entropy 

entropy of a system 

scalar temperature 

absolute temperature 

reduced temperature 

specific internal energy 

specific volume 

relative volume 

volume of a system 

reduced volume 

work 

mole fraction of a gas in a mixture of gases 
compressibility factor 

compressibility factor at the critical point 

a function of the temperature 

* a property of the perfect gas 

at the critical point 
of the mixture 
constant pressure 
constant entropy 
constant temperature 
constant volume 




CHAPTER 19 


THERMODYNAMICS OF COMBUSTION 

19:1. Introduction. All of the power cycles that have been discussed 
in earlier chapters presume that a reservoir is available from which heat 
may be continuously withdrawn at a temperature above that of the atmos¬ 
phere. If this reservoir is finite in extent and is not at the same time 
receiving energy at a rate at least equal to that at which it supplies heat 
to the working fluid of the heat engine, its temperature must ultimately 
decline to that of the atmosphere and the manufacture of power will cease. 

The engineer provides for the continuous replenishment of the energy 
of the source by means of the combustion of some sort of fuel. Fuels may 
be solids (as coal), liquids (oil or gasoline), or may be in the form of a 
gas (for example, natural gas), but all have the common characteristic of 
containing at least some proportion of certain elements in a form such 
that, if they are mixed with air and ignited by means of a spark or flame, 
they will unite with the oxygen of the air to form new substances, at the 
same time releasing energy of which the engineer can make use. The 
process by means of which this energy is released is called combustion. 
The combination of the fuel with the air which is to supply the oxygen for 
combustion is given the name of the mixture, or the reactants, and the sub¬ 
stance, or combination of substances, that results from the combustion 
process is termed the products. 

The combustible elements of a fuel that are of importance to the 
engineer include carbon (C), hydrogen (H), and, of much less importance, 
sulfur (S). The combustion of carbon may produce carbon monoxide 
(CO) or carbon dioxide (C0 2 ), according to the conditions that exist in the 
course of the reaction. If carbon monoxide is formed, the amount of 
energy released will be less than if the products consist of carbon dioxide 
but the CO is capable of a further reaction with additional amounts of 
oxygen that will produce C0 2 as the end product. Further increments of 
energy equal in amount (if the initial and final temperatures and pres¬ 
sures are the same in both cases) to the original deficit will be released; 
this is a requirement of the First Law. Hydrogen burns to form water 
(H 2 0) and sulfur forms S0 2 as a result of the engineering combustion 
process. 

The reason for the energy release that accompanies combustion is found 
in the fact that a given mass of the products stores a much smaller amount 

458 


THERMODYNAMICS OF COMBUSTION 


459 


of energy than an equal mass of the reactants at the same pressure and 
temperature. When combustion occurs, this excess of energy may have 
a number of effects. If the space in which combustion takes place is sur¬ 
rounded by rigid walls and is insulated against heat flow, its passage as 
heat or work to the surroundings is prevented and it must remain as stored 
energy in the products. But this means that the products, in order to 
store the same amount of energy as the reactants originally possessed, 
must rise in temperature above the temperature of the mixture as the 
reaction began; at this elevated temperature the products provide a source 
from which heat may be withdrawn at'high temperature for the purposes 
of the power cycle. If enough heat leaves the products, their temperature 
may be returned to the original temperature of the reactants; this heat 
flow then measures the amount of energy release and is called the heat of 
combustion at the temperature at which the reaction began and ended. 
Because the relative energy-storage capacities of the reactants and the 
products may differ at different temperatures, a corresponding variation 
of the heat of combustion according to the base temperature that is 
selected is to be expected. 

Based upon our study of the Carnot cycle, we observe that not all of 
the heat of combustion may be changed to the form of work. The pro¬ 
portion that is available for that purpose will depend upon the tempera¬ 
ture at which the heat of combustion can be delivered for the purposes of 
the cycle. As the result of combustion, the temperature of the products 
rises to a certain maximum which will depend upon the heat of combus¬ 
tion, the mass of the products, and their specific heat. The first incre¬ 
ment of heat may be withdrawn to supply the power cycle at this maxi¬ 
mum temperature and may be used with maximum effectiveness in the 
manufacture of work, although even this increment is not, as we have 
seen, possible of complete cyclic conversion into the more valuable form 
of energy. As additional amounts of heat are withdrawn, the tempera¬ 
ture of the products declines and the effective source temperature with it, 
with a corresponding decrease in the efficiency of conversion into work. 
Assuming that combustion began at the temperature of the atmosphere, 
the last increments of the heat of combustion are almost entirely ineffec¬ 
tive for the purpose of work delivery by the power cycle, for they can be 
delivered at temperatures only insignificantly above the temperature of 
the atmosphere. 

A larger proportion of the energy released as the result of combustion 
can be converted into work if the power cycle is supplied with heat at an 
essentially constant temperature which is as high as is possible and prac¬ 
tical. This implies that the combustion process shall be continuous, with 
a steady stream of fuel uniting with a stream of air to produce a stream of 
products. Heat is removed from these products for the purposes of the 


460 


BASIC ENGINEERING THERMODYNAMICS 


power cycle, but their final temperature remains at a level such that the 
heat engine is not handicapped in attaining a reasonable efficiency in con¬ 
verting this heat into work. Since the products are not returned to the 
original temperature of the reactants, a part of the heat of combustion 
remains in those products as stored energy after they have contributed 
their dividend of heat to the power cycle. 

The temperature at which heat can be delivered for the purposes of the 
power cycle is, as we have seen, important in limiting the efficiency of the 
heat engine; that temperature cannot exceed the maximum temperature 
that can result from combustion and thus, as stated above, depends not 
only upon the heat of combustion but also upon the mass and the specific 
heat of the products. The engineer obtains the oxygen for combustion 
from air, a gas which is a mixture of about one-fifth oxygen with four- 
fifths nitrogen and other inert gases. These inert gases form a part of the 
products as well as of the reactants and increase the mass of those prod¬ 
ucts; this increased mass lowers the maximum temperature attainable. 
Air in excess of the minimum amount necessary to supply the oxygen 
required for combustion is often used by the engineer in order to ensure 
that each particle of fuel will have in its immediate vicinity enough oxygen 
to supply its needs for complete combustion; the use of excess air still fur¬ 
ther increases the weight of products per unit mass of fuel burned and 
limits the maximum temperature. 

The heat of reaction is based on the assumption that combustion is 
complete. The maximum temperature that can be reached as the result 
of combustion is limited not alone by the possibility that the mixture of 
reactants is not homogeneous and that some particles of fuel may be 
starved for oxygen and remain unburned while other particles have an 
oversupply. In addition, as we shall see later, the reaction is, in limited 
degree, reversible. Thus if carbon dioxide, for example, is raised to high 
temperature, certain proportions will break down into the elements of 
carbon and oxygen. When the reaction takes place in this direction, it 
is called dissociation; according to the First Law, the heat of reaction must 
be negative in dissociation. Depending upon the temperature and pres¬ 
sure, there are certain fixed proportions of the compound and its elements 
present in a mixture that is in completely stable equilibrium, and this 
factor also limits the maximum temperature of the products to a level 
below that based on the assumption of complete combustion. 

The combustion of the engineer is a controlled combustion and takes 
place within a confined space called the combustion space. The products 
of combustion come in contact with the walls which bound this space. 
The materials of which these walls are built are limited in the tempera¬ 
tures which they can withstand over long periods of time, and this often 


THERMODYNAMICS OF COMBUSTION 


461 


places a practical limit on the temperature at which the products may be 
allowed to supply heat to the working fluid of the heat engine. 

The residual energy remaining in the products after they have made 
their direct contribution of heat for the purposes of the heat engine may be 
partly used. If the temperature of the reactants, instead of being the 
temperature of the atmosphere, had been elevated above that tempera¬ 
ture, the maximum temperature that would have resulted from combus¬ 
tion would have been correspondingly, though not equally, increased. A 
larger fraction of the heat of combustion is therefore deliverable to the 
heat engine. By means of suitable heat exchangers (called air preheaters) 
the air for combustion can receive heat from the products after the latter 
have completed their commitments in the supply of heat to the power 
cycle and can be raised in temperature as the temperature of the products 
falls. Or it is often the case that all of the heat supplied the power cycle 
need not be furnished at the same high level of temperature. An example 
is found in the Rankine cycle, in which a part of the heat supplied is used 
to heat the liquid from the temperature of the refrigerator to the tempera¬ 
ture at which it is evaporated. At least a part of this heat may be sup¬ 
plied to the liquid as it passes through a heat exchanger, called an econo¬ 
mizer, in which it receives heat from the products after they have made 
their major contribution of heat to the cycle but before they are discarded 
to the atmosphere. In theory, the temperature of the products could, by 
methods such as these, be returned to approximately the temperature of 
the reactants, and practically all of the heat of combustion could be 
utilized. In practice this is not feasible, for it would require large and 
expensive heat exchangers and would discharge the products in a state 
such that some of the water vapor which they usually contain would con¬ 
dense and cause corrosion. Corrosion would be especially likely if the 
fuel contained sulfur and the products sulfur dioxide, since the combina¬ 
tion of liquid water and sulfur dioxide would produce highly corrosive 
sulfurous acid. The minimum practical temperature of the products as 
they are finally discharged to the atmosphere is therefore about 300°F, 
although temperatures above that level are much more common. 

Much has been said in preceding chapters about the cycle of the work¬ 
ing fluid. In the Rankine cycle, for example, this cycle is carried out in 
the boiler, the prime mover, the condenser, and the liquid pump. But the 
real vapor power plant must also include a furnace to supply the heat that 
is added to the working fluid as it passes through the boiler. The sub¬ 
stances that pass continuously through this furnace constitute a second 
system, which does not follow a cyclic pattern since the products of com¬ 
bustion are never, at least under the guidance and control of the engineer, 
restored to the form of the reactants that entered the process. In the 


462 


BASIC ENGINEERING THERMODYNAMICS 


internal-combustion power plant the reactants, and later the products, 
constitute the working fluid and a cycle of the working fluid is not 
traversed. In order to study the operation of the internal-combustion 
engine from the standpoint of a closed cycle, it was necessary in Chap. 11 
to introduce the concept of the air-standard cycle. A more realistic 
approach to the problem consists in analyzing the individual processes 
through which the fluid progresses from the time it enters the plant in the 
form of the reactants until it leaves as the products of a combustion that 
has taken place in the interim. Differences in chemical composition and 
in the mass that is included within the boundaries of the system under 
study must be considered in making this type of analysis. Some of these 
processes, as when the reactants enter or the products leave, are flow 
processes and may be analyzed as such, although truly steady-flow condi¬ 
tions are not always present. Others are typically nonflow in character, 
and the equations of Chap. 2 will apply, but with the reservation that the 
system is not a simple system. In either case, E replaces U and must be 
calculated with respect to the chemical composition of the fluid at the end 
points of each process. 

19:2. The Combustion Equation. The combustible elements of 

importance to the engineer are carbon, hydrogen, and sulfur. To support 
combustion, he must also arrange for a supply of oxygen. The combus¬ 
tion equation provides a key to the relative quantities of the combustible 
and the oxygen that enter the reaction and the products that result from 
it. Thus, in the complete combustion of carbon to form carbon dioxide, 
we may write 

C ~b O 2 —* CO 2 (19:1) 

This equation may be interpreted in a number of ways. For example, it 
may be read as “ 1 molecule of carbon unites with 1 molecule of oxygen to 
form 1 molecule of carbon dioxide.’’ We have already learned (Chap. 9) 
that a mole is a standard number of molecules, and the magnitude of the 
reaction as originally described can be multiplied by this number; the 
equation can correspondingly be read as describing the union of 1 mole of 
carbon with 1 mole of oxygen to form 1 mole of carbon dioxide. It will 
be observed that it is not necessary that the number of moles of the prod¬ 
ucts shall equal the number of moles of the reactants. On the other hand, 
it is required that a mass balance shall exist on the two sides of the equa¬ 
tion. The weight of the standard number of molecules that is described 
as 1 lb-mole is numerically equal, in pounds, to the molecular weight m. 
Thus Eq. (19:1) may be interpreted as describing the reactants as con¬ 
sisting of 12 lb of carbon, 32 lb of oxygen, while the weight of C0 2 that 
constitutes the products is 44 lb. If all of the substances that enter into 
the reaction and result from it are gases and their pressure is low enough 


THERMODYNAMICS OF COMBUSTION 


463 


so that they can be considered to be perfect gases, their volumes, at the 
same pressure and temperature, will be in proportion to the relative num¬ 
ber of moles of each that takes part in the reaction. Thus, if the carbon 
reactant in Eq. (19:1) is a gas, we may interpret the equation as represent¬ 
ing the union of 1 volume of carbon with 1 volume of oxygen to form 1 
volume of carbon dioxide; the inference is, of course, that these volumes 
are being measured at the same pressure and temperature, i.e., that the 
products are at the same pressure and temperature as the reactants. On 
the other hand, it is quite possible that the carbon may be in the solid or 
the liquid phase as it enters the reaction, in which case its volume would 
be negligible as compared with the molal volume of the gaseous oxygen 
and carbon dioxide; if this qualification were made, we should be forced 
to alter our statement to indicate that the volume of products equals the 
volume of the reactants for this particular reaction. Thus, if the com¬ 
bustion equation is to be read in terms of the relative volumes of the 
reactants and the products, the phase of the fuel as it enters the combus¬ 
tion must be specified. This specification is often made by using the 
parentheses-enclosed letters (s), (1), or (g), meaning, respectively, solid, 
liquid, and gas, after each of the reactants and the products to indicate 
their phase and to guide the reader in estimating relative volumes. 

Let us review the various interpretations of Eq. (19:1). 

Reactants Products 

C(g) + 0 2 (g) -> C0 2 (g) (19:1a) 

1 mole C + 1 mole 0 2 —> 1 mole C0 2 (19:16) 

12 lb C + 32 lb 0 2 = 44 lb C0 2 (19:1c) 

1 volume C + 1 volume 0 2 —> 1 volume C0 2 (19:1c?) 

Of these equations [note, however, that only (19:1c) is, of necessity, an 
equality] 19:1 d is an approximation which is based on the assumption that 
the reactants and the products are perfect gases and that the volume of 
reactants and products is measured at the same pressure and temperature. 
If (19:1a) had been written 

C(s) + 0 2 (g) -> C0 2 (g) 

(19:lrf) would have undergone a change to the form 

0 volume C + 1 volume 0 2 —» 1 volume C0 2 

and the volume of reactants and products would have been the same. An 
additional approximation has thus been introduced in considering the 
volume of the solid to be negligible. 

In addition to Eq. (19:1), the basic combustion equations with which 
the engineer is concerned are 


464 


BASIC ENGINEERING THERMODYNAMICS 


C -T ^0 2 — 

CO 

(19:2) 

CO +^0 2 — 

-> co 2 

(19:3) 

H 2 + 40 2 — 

+ h 2 o 

(19:4) 

S + 0 2 — 

-> so 2 

(19:5) 


For each of these basic reactions, a set of expressions analogous to 
(19:1a) to (19:1c?) may be framed from which the relative number of 
moles, the weights, and the volumes of reactants and products may be 
obtained and compared when pure oxygen is supplied for the purposes ol 
combustion. But the engineer furnishes oxygen in the dilute form of air, 
which is essentially a mixture of 0.21 mole of oxygen per 0.79 mole of inert 
nitrogen, and the mass and volume of both reactants and products are 
correspondingly increased. Per mole of oxygen entering the reaction, 
-£f, or 3.76, moles of nitrogen are also introduced and will appear in the 
products, since this nitrogen takes no active part in the chemical change. 
Thus when exactly enough air is furnished to supply the oxygen necessary 
to burn the carbon to C0 2 , Eq. (19:1) becomes 

C + 0 2 + 3.76N 2 C0 2 + 3.76N 2 (19:6) 

and calculation indicates that the proportion of the moles of products to 
the moles of reactants is 4.76:5.76 while the weight balance is as follows: 

12 lb C + 32 lb 0 2 + 105.4 lb N 2 = 44 lb C0 2 + 105.4 lb N 2 

The fuels with which the engineer deals are almost invariably more 
complicated combinations of combustible elements than are contemplated 
in the basic combustion equations (19:1) to (19:5). They may also, as in 
coal, contain considerable percentages of noncombustible substances. 
Let us see how the basic equations may be applied in writing the com¬ 
bustion equation for butane (C 4 Hi 0 ), for example. It will be assumed 
that combustion is complete (C to C0 2 ); the products will therefore con¬ 
sist of C0 2 and H 2 0, and we may write the combustion equation in unbal¬ 
anced form (showing each of the reactants and each of the products but 
not the number of molecules of each) as 

C 4 H 10 T O 2 —* C0 2 -T H 2 0 

Noting that the same number of carbon atoms must appear in both react¬ 
ants and products, we see that the number of molecules of C0 2 must be 
4. In a similar manner, based on a balance of the number of atoms of 
hydrogen, the number of molecules of H 2 0 that will appear in the products 
is evidently 5. The number of atoms of oxygen in 4 molecules of C0 2 is 
8 , and the number of oxygen atoms in 5 molecules of H 2 0 is 5, a total of 
13 atoms of oxygen; to balance the equation requires that 6 i molecules be 


THERMODYNAMICS OF COMBUSTION 465 

included in the reactants. The balanced equation may then be written as 

-C 4 H 10 + 6^0 2 —> 4 CO 2 + 5H 2 0 (19:7) 

The equation may now be read in terms of moles rather than molecules. 
If the oxygen is furnished in air, it will be necessary to add (6£)(3.76) = 
24.44 moles of nitrogen to both reactants and products in Eq. (19:7), or 

C 4 H 10 + 6^0 2 + 24.44N2—* 4 CO 2 + 5H 2 0 + 24.44N2 (19:8) 

The weight balance for this equation may now be written as 

58.1 lb C 4 Hio + 208 lb 0 2 + 685 lb N 2 = 176 lb C0 2 + 90.1 lb H 2 0 

+ 685 lb N 2 

The air-fuel ratio is often a matter of interest to the engineer; this is the 
ratio of the mass of air supplied for combustion to the mass of fuel. For 
the combustion that is represented by Eq. (19:8), the mass of air supplied 
to burn 1 mole, or 58.1 lb, of fuel is the sum of the masses of the oxygen 
and the nitrogen in the reactants, or 893 lb; the air-fuel ratio is therefore 
893/58.1 = 15.4 lb air per pound C 4 Hi 0 . 

The value of 15.4 lb of air per pound of C 4 Hi 0 which is derived as the 
air-fuel ratio above is based on supplying the minimum amount of air that 
could furnish the oxygen necessary for complete combustion; this is called 
the theoretical air. 

The engineer cannot obtain a mixture of fuel and air such that sufficient 
oxygen will be available to each particle of fuel for its complete combustion 
without providing what he calls excess air , i.e ., air in excess of the theo¬ 
retical requirements. He expresses the amount of this excess air as a 
percentage based upon the theoretical. Thus if 10 per cent excess air is 
supplied in the combustion of C 4 Hi 0 , the weight of air supplied must be 
( 1 . 1 ) (893) = 982 1b per 58.1 lb of C 4 Hi 0 , corresponding to an air-fuel 
ratio of 982/58.1 = 16.9 lb of air per pound of C 4 Hi 0 . The extra oxygen 
and nitrogen will appear in unchanged form in the products. Based upon 
10 per cent excess air, Eq. (19:8) would assume the form 

C 4 H 10 + 7.150 2 + 26.88N 2 —> 4C0 2 + 5H 2 0 + 0.65O 2 + 26.88N 2 

On the other hand, insufficient air may be provided for complete com¬ 
bustion. If the deficit is small, it will, in the combustion of a hydrocar¬ 
bon, probably result in the presence of a proportion of CO in the products; 
if large, free carbon, or even free hydrogen, may appear. As an example, 
let us assume that 90 per cent of the theoretical air is supplied in burning 
C 4 H 10 and that the products consist only of C0 2 , CO, H 2 0 , and N 2 . We 
may then write the equation in the form 

C 4 H 10 + (0.9)(6.5)O 2 + (0.9) (24.44)N 2 —> aC0 2 + bCO + 5H 2 0 + 22 N 2 


466 


BASIC ENGINEERING THERMODYNAMICS 


in which a and b are, respectively, the number of molecules of CO 2 and CO 
in the products per molecule of C 4 H 10 in the reactants. Based on the 
necessary balance of carbon atoms, 

a + b = 4 

A similar balance of the number of oxygen atoms requires that 

2a + b + 5 = (2) (0.9) (6.5) = 11.7 

Solving these two equations simultaneously, a = 2.7 and 6 = 1.3 The 
balanced equation is 

C 4 H 10 + 5.850 2 + 22N 2 —* 2.7C0 2 + 1.3CO + 5H 2 0 + 22N 2 

Example 19:2. The fuel is coal with an ultimate analysis 1 as follows, 

C, 78; H, 3; O, 2; S, 1; N, 1; H 2 0, 4; ash, 11 

in which C, H, O, S, and N are the corresponding elements, H 2 0 is the surface mois¬ 
ture, and the ash is a mixture of noncombustible substances. The figures give the 
percentages by mass of each constituent. Write the combustion equation for 100 lb 
of this coal, and calculate the air-fuel ratio ( a ) when complete combustion is assumed 
with theoretical air supplied, (6) when complete combustion is assumed with 50 per 
cent excess air, and (c) when the air supplied is 90 per cent of the theoretical but the 
hydrogen and sulfur burn completely and there is no free oxygen in the products. 

Solution. The weight of 100 lb is an artificial mole unit often used by the engineer 
in analyzing the combustion of fuels which, like coal, are described in terms of their 
ultimate analyses. Thus 1 “mole” of this coal contains 78 lb of carbon, 3 lb of hydro¬ 
gen (in addition to the hydrogen in the surface moisture), etc. Dividing these weights 
by the atomic weights, an equivalent chemical formula for the coal (exclusive of the 
ash, which is noncombustible and does not take part in, or change its phase as the 
result of, the reaction) is 

C 7 |H ? O i 2 b S 3 i 5 N i i j (s) + (H 2 0 )j 4 8 ( 1 ) = Ce. 5 H 3 O 0 . 12 S 0 . 03 N 0.07 (s) + 0.22H 2 O(l) 

(a) Writing the unbalanced equation, 

C 6 . 5 H 3 O 0 . 12 S 0 . 03 N 0.07 4" 0.22H 2 O 4~ 0 2 N 2 —■» C0 2 4~ H 2 0 4- S0 2 4- N 2 

The number of hydrogen atoms in the reactants is 3 4- (2) (0.22) = 3.44. The num¬ 
ber of moles of H 2 0 in the products must therefore be 3.44/2 = 1.72. The number of 
moles of C0 2 in the products must be equal to the number of carbon atoms in the 
reactants, or 6.5. Similarly, based on a balance of a sulfur atoms, the number of 
moles of S0 2 in the products is 0.03. 

Placing these coefficients on the right side of the equation, the number of atoms of 
oxygen in the products is calculated as (6.5) (2) 4- 1.72 4- (0.03) (2) = 14.78. Sub¬ 
tracting the number of atoms of oxygen in the fuel, the number of moles of 0 2 in the 
reactants is calculated as (14.78 — 0.12 - 0.22)/2 = 7.22. The number of moles of 

1 The ultimate analysis states the percentages of the masses of the various constitu¬ 
ents, i.e., is expressed on a gravimetric basis. It is usually employed in describing 
solid or liquid fuels. A volumetric analysis is often used when the fuel is a gas and 
may be converted to the gravimetric basis by applying the methods of Chap. 10. 


THERMODYNAMICS OF COMBUSTION 467 

nitrogen in both reactants and products is 0.07/2 + (3.76) (7.22) = 27.18. The 
balanced equation may now be written as 

C6.6H 3 Oo.i2So.o3No.o7(s) + 0.22H 2 O(l) + 7.220 2 (g) + 27.15N 2 (g) -> 6.5C0 2 (g) 

+ 1.72H 2 0 + 0.03S0 2 (g) + 27.18N 2 (g) 
The air-fuel ratio is (7-22) (32) +^(27.15) (28) = ggi , b air per pound coaI 

(6) The use of 50 per cent excess air will increase the number of moles of 0 2 in the 
reactants by (0.5) (7.22) = 3.61, making a total of 10.83 moles. The extra 3.61 
moles will appear as 0 2 in the products. The number of moles of N 2 in both reactants 
and products will be increased in the same proportion, or by 13.58 moles. The com¬ 
bustion equation is 

C 6 . 5 H 3 O 0 . 12 S 0 . 03 N 0.07 + 0.22H 2 O + 10.83O 2 + 40.73N 2 -* 6.5C0 2 + 1.72H 2 0 

+ 0.03S0 2 + 3.610 2 + 40.76N 2 

The air-fuel ratio is (1.5) (9.91) = 14.87 lb of air per pound of coal. 

(c) The number of moles of 0 2 in the reactants is (0.9) (7.22) = 6.50. The number of 
moles of N 2 is, similarly, 24.43. Writing the combustion equation in partially bal¬ 
anced form, 

C 6 . 5 H 3 O 0 . 12 S 0 . 03 N 0.07 4- 0.22H 2 0 -f- 6.50O 2 -(- 24.43N 2 —■» aC0 2 + 6CO + 1.72H 2 0 

4- 0.03S0 2 + 24.46N 2 

where a and b are, respectively, the unknown number of moles of C0 2 and CO that 
are formed as the result of the combustion of 100 lb of coal. A balance based on the 
number of carbon atoms shows that 

a + b = 6.5 

Balancing the number of oxygen atoms, we may write 

0.12 + 0.22 + (6.5) (2) = 2a + b + 1.72 + (0.03) (2) 
or 

2 a + b = 11.56 

Solving simultaneously, a = 5.06 and b = 1.44. The combustion equation becomes 

C 6 . 5 H 3 O 0 . 12 S 0 . 03 N 0.07 + 0.22H 2 O + 6.5002 + 24.43N 2 —> 5.06CO 2 + 1.44CO 

+ 1.72H 2 0 + 0.03S0 2 + 24.46N 2 

The air-fuel ratio is (0.9) (9.91) = 8.92 lb of air per pound of coal. 

19 : 3. Analysis of Products. In setting up the combustion equations 
of the preceding article, it has been necessary to make certain assump¬ 
tions. Chiefly, these assumptions have concerned the substances of 
which the products are composed. For example, in analyzing the com¬ 
bustion of C 4 H 10 with insufficient air for complete combustion, it has been 
assumed that all of the oxygen entered into some sort of combustion with 
the combustible elements of the fuel. Yet it is entirely possible that, 
owing to incomplete mixing of fuel and air, this would not be the case and 
some free oxygen would be present in the products. Similarly, even when 
excess air is provided, the products may include carbon monoxide. The 



468 


BASIC ENGINEERING THERMODYNAMICS 


reaction equation can be accurately written only with a knowledge of the 
composition of the products, and without this information the engineer is 
handicapped in taking steps toward the improvement of combustion 
conditions in order to obtain the maximum from his fuel. 

An incomplete analysis of the products is usually satisfactory for the 
engineer’s purposes if he has a fairly accurate knowledge of the composi¬ 
tion of the fuel. This analysis of the products of combustion (usually 
called a flue-gas analysis) is customarily made on a volumetric basis, using 
the Orsat apparatus. The method consists in withdrawing some of the 
gas and segregating a sample of standard volume, as measured at atmos¬ 
pheric pressure and temperature, in a burette designed for that purpose. 
This sample is then passed through a solution of potassium hydroxide 
that will absorb C0 2 and returned to the measuring burette, where, again 
at atmospheric temperature and pressure, the remaining volume is noted; 
the volumetric percentage of CO 2 is determined by difference. The free 
oxygen is next removed by passing what remains of the original sample 
through a solution of pyrogallic acid; the remaining volume is again 
measured to obtain, by difference, the volumetric percentage of oxygen. 
Finally, a solution of cuprous chloride removes the CO from the part of 
the sample still remaining and makes possible the determination of the 
volumetric percentage of that constituent. The remaining volume is 
assumed to represent the percentage of nitrogen. The sample also con¬ 
tains water vapor which is not removed, but the partial pressure of the 
water vapor remains constant since the conditions under which the volume 
is measured are such as to ensure that it will be saturated with water 
vapor and since the temperature remains constant throughout the analy¬ 
sis. Thus the volumetric proportions of C0 2 , 0 2 , CO, and “nitrogen” 
are obtained on the dry basis, i.e., as if the water vapor had been removed 
from the sample prior to its analysis. Additional solutions for the removal 
of other gases may be added if a more detailed analysis is desired; some¬ 
times an analysis for C0 2 alone is made. 

As an example of a way in which the engineer may make use of an 
analysis of the products of combustion, let it be supposed that in the burn¬ 
ing of butane the following Orsat analysis (volumetric, dry basis) is 
obtained: 


C0 2 , 10.0%; 0 2 , 5.0%; CO, 0.5%; N 2 (by difference), 84.5% 

The air-fuel ratio and the per cent of excess air are required. 

The combustion equation may be written as 

C,Hi„ + a0 2 + 3.76aN 2 -> bC0 2 + ^ CO + ^ 0 2 + cH 2 0 


+ 3.76<jN 2 




THERMODYNAMICS OF COMBUSTION 


469 


in which a is the number of moles of oxygen supplied, b is the number of 
moles of CO2, and c is the number of moles of H 2 0 in the products per 
mole ot C4H10 burned. The number of moles of nitrogen, it will be noted, 
bears a fixed ratio to the number of moles of oxygen which is based on 
their relative proportions in air; the number of moles of nitrogen in 
reactants and products must be the same. The number of moles of CO 
and of 0 2 in the products similarly bear fixed ratios to the number of 
moles of CO2, ratios which are based on the relative percentages as shown 
by the Orsat analysis. 

From a hydrogen balance, 

c = 5 

From a carbon balance, 

7 , 0.56 . 7 o 07 

b + loo = 4 or 6 = 381 

From an oxygen balance, 

n 56 5 0 

2a = 26 + j q - q + ^ q 26 + 5 or a = 8.31 

Substituting these values in the combustion equation, 

C 4 H 10 + 8.3102 + 31.2N 2 -* 3.8ICO2 + 0.19CO + 5H 2 0 + 31.2N 2 

The weight of air supplied per mole of C4H10 is (8.31) (32) + (31.2) (28) = 
1140 lb. The air-fuel ratio = 1140/58.1 = 19.6 lb per pound C4H10. 
This corresponds to 19.6/15.4 — 1 = 0.27, or 27 per cent excess air. 

The data may now be checked for a reasonable degree of consistency. 
It will be noted that the number of moles of nitrogen in the products 
might also have been expressed in a proportion based on the Orsat 
analysis as 84.56/10.0 = (8.45) (3.81) = 32.2 moles. This checks the 
number as based on the oxygen supplied within approximately 3 per cent, 
about as close agreement as can be expected. 

Example 19:3. In burning the coal of Example 19:2, an Orsat analysis shows the 
following percentages by volume for the gaseous products: CO 2 , 12 per cent; 0 2 , 
7 per cent; CO, 0.5 per cent; N 2 (by difference), 80.5 per cent. An analysis of the 
cinder in the ash pit shows it to be 30 per cent unburned carbon. Write the combus¬ 
tion equation, and determine the percentage of excess air. 

Solution. The weight of ash is 11 lb per “mole” of coal. If the cinder is 30 per 
cent unburned carbon, the total weight of cinder is 11/0.7 = 15.7 lb per 100 lb of coal, 
and 4.7 lb (4.7/12 mole) of carbon remains unburned. A partially balanced com¬ 
bustion equation may be written. 

C 6 . 5 H 3 O 0 . i 2 So. 03 N 0.07 + 0.22H 2 O + a0 2 + 3.76 aN 2 —* 6C0 2 + cCO + d0 2 

+ 0.39C + 1.72H 2 0 + 0.03S0 2 + eN 2 

Based on a balance of carbon atoms, 


b + c = 6.5 - 0.39 = 6.11 


( 1 ) 




470 


BASIC ENGINEERING THERMODYNAMICS 


From an oxygen balance, 

0.12 + 0.22 + 2a = 26 + c + 2d + 1.72 + 0.06 


or 


2a - 26 - c - 2d = 1.44 

(2) 

From the Orsat analysis, 


c = ^ b = 0.04176 

1 z 

(3) 

and 


d = j^b = 0.5836 

(4) 


From Eqs. (1), (3), and (4), 

6 = 5.87; c = 0.24; d = 3.42 
Substituting these values in (2), 
a = 10.13 

Making a nitrogen balance, 

0,07 + (3,76) (10.13) (2) = 3g 1Q 
2 

The combustion equation may now be written: 

C 6 . 5 H 3 O 0 . 12 S 0 . 03 N 0.07 + 0.22H 2 O + 10.1302 + 38.08N 2 —> 5.87C0 2 + 0.24CO 

+ 3.420 2 + 0.39C -b 1.72H>0 + 0.03S0 2 + 38.10N 2 

From the solution to Example 19:2(a), the number of moles of oxygen in the theoreti¬ 
cal air is 7.22. In the present example, therefore 

„ . 10.13-7.22 _ . no . nQ 

Excess air = -- = 0.403, or 40.3 per cent 

«.zz 

Checking the combustion equation, it is noted that the number of moles in the prod¬ 
ucts is 38.10. From the Orsat analysis, the ratio e/b should be in the proportion 
80.5/12 = 6.7, or e = 6.7 b = (6.7) (5.87) = 39.4. This would be considered in 
fairly close agreement. 

19 : 4. The heat of combustion is a measure of the heat that may be 
liberated as a result of the combustion reaction if the products are returned 
to the original temperature of the reactants. This heat flow, when posi¬ 
tive, is away from the system which undergoes the reaction; the conven¬ 
tion as to sign in chemistry is thus opposite to that which we have used in 
thermodynamics, and this must be kept in mind as the basic thermo¬ 
dynamic relations are applied. 

The heat of combustion of solids and liquids is usually measured experi¬ 
mentally by means of a bomb calorimeter. This is a rigid-walled vessel 
in which a carefully measured sample of the fuel is placed, together with 
oxygen in an amount more than sufficient to ensure complete combustion. 




THERMODYNAMICS OF COMBUSTION 


471 


The vessel, after being closed, is placed in a water bath, and the fuel is 
ignited by means of an electric current. As the fuel burns, the tempera¬ 
ture of the bomb and the water will rise only moderately because of the 
high capacity for energy storage of the water bath; although the final 
temperature is somewhat above the initial temperature of the reactants, 
allowance may be made for this and other factors in calculating the heat 
of combustion. 

In the bomb-calorimeter process, the system is a closed system, and 
the combustion takes place at constant volume; the heat of combustion as 
determined by this method is called the constant-volume heat of combustion. 
No work accompanies the reaction. Taking into consideration the 
reversal of the sign convention to which attention has been directed above, 
Eq. (2:2) may be applied to write 

Qv — lQ 2 = [El — E 2 ]t = [^mixture -^products] T (19:9) 

in which Q v is the constant-volume heat of combustion, the subscripts 1 
and 2 refer, respectively, to the beginning and to the end of the combus¬ 
tion process, and E (for total stored energy) is used since this is not a sim¬ 
ple system. The subscript T indicates that the values of E must be based 
on the same (base) temperature. We may substitute £/ mixture and U pioduct8 , 
respectively, for E mixture and E producta in Eq. (19:9) if U for the products not 
only is measured at the same temperature at which the internal energy of 
the mixture is measured but also is based on the same reference level of 
total stored energy. 1 On this basis the heat of combustion at constant 
volume, obtained experimentally, measures the difference between the 
internal energy of the mixture and the internal energy of the products at 
the same temperature, and we may write 

Q v = [ U mixture — ^products] T = [~AU]t (19 TO) 

The fuel tested in the bomb calorimeter is characteristically a solid or a 
liquid and thus occupies negligible volume. Also, when H 2 0 is among the 
products, almost all will appear as a liquid instead of a gas in those prod¬ 
ucts because of the low final temperature and the limited volume of the 
calorimeter. Thus the combustion equation for butane, for example, as 
that combustion is carried out in the bomb calorimeter, would appear as 

C4Hio( 1) + 6^0 2 (g) —> 4C0 2 (g) + 5H 2 0(1) 

The value of the heat of combustion that is obtained in the use of the 
bomb calorimeter is based on these specifications as to the phase of the 
various substances that enter into and result from the reaction. How¬ 
ever, if it is desired to obtain the constant-volume heat of combustion 

1 Thus U would include chemical energy. 


472 


BASIC ENGINEERING THERMODYNAMICS 


that corresponds to the combustion equation 

C 4 H 10 (g) + 6i0 2 (g) -► 4C0 2 (g) + 5H 2 0(g) 

all that is necessary is to add to the bomb-calorimeter value the internal 
energy of vaporization of C4H10 and subtract the internal energy of 
vaporization of H 2 0, both at the base temperature to which the reaction 
is referred. 

Returning to Eq. (19:10), the rate of change of internal energy with 
temperature will usually differ somewhat for the products from the similar 
rate of change for the mixture, indicating that the difference —A U of that 
equation will depend upon the base temperature at which the reaction is 
begun and concluded. Thus the heat of combustion must be stated with 
reference to this base temperature; the standard base temperature is 
25°C, equivalent to 77°F. A change to any other base temperature may 
be made if the specific heats of the mixture and of the products are known, 
so that the changes of internal energy of mixture and of products over the 
temperature interval may be calculated and compared. 

The heat of combustion of gases is usually obtained experimentally by 
the use of the steady-flow gas calorimeter. In this device the steadily 
flowing stream of fuel is mixed with a stream of air that carries more than 
sufficient oxygen to ensure complete combustion and is burned continu¬ 
ously. The fuel supply is measured volumetrically by means of a gas 
meter. The products are cooled to the initial temperature of the reactants 
by means of a water jacket. The rise in temperature of the water passing 
.through this jacket is a key to the constant-pressure heat of combustion , 
since the reaction takes place at constant pressure. 

The gas calorimeter is a steady-flow device and the system is an open 
system; Eq. (3:5) may be applied, giving 

Qp 1Q2 \H m ixture H prodxicta ] TiP [ A H\x t p (19:11) 

in which Q p is the constant-pressure heat of combustion and the subscripts 
1 and 2 again refer to states at the beginning and at the end of the reac¬ 
tion. H ixture and H product8 are measured at the same pressure and tem¬ 
perature, as is indicated by the subscripts T and p, and are based on the 
same total weight and the same reference state (U includes chemical 
stored energy). The phase in which each of the substances that compose 
the mixture and the products appears will affect the constant-pressure 
heat of combustion as it did the heat of combustion at constant volume, 
but again an allowance may be made for a difference in this respect, this 
time by means of an adjustment based on the enthalpy of vaporization 
instead of the internal energy of vaporization, as when adjusting the 
constant-volume heat of combustion to differences in phase. 


THERMODYNAMICS OF COMBUSTION 


473 


Table 19:1. Constant-pressure Heats of Combustion* 

At 77°F 


Basic combustion equations 


Fuel 

Formula 

Phase 
of fuel 

Combustion equation 

-AH, 
Btu/ 
lb fuel 


Hydrogen. 

h 2 

Gas 

H 2 (g) + K>2(g) -> H.O(l) 

60,958 


Carbon. 

C 

Solid 

C(s) + 0 2 (g) 

- C0 2 (g) 

14,087 


Carbon monoxide 

CO 

Gas 

CO(g) + ■g-0 2 (g) —■> C0 2 (g) 

4,344 


Sulfur. 

s 

Solid 

S(s) + 0 2 (g) 

-> so 2 (g) 

3,940 




Hydrocarbons 







Phase of products 

AH 

hf g of fuel 



Phase 




at 77°F, 

Fuel 

Formula 

of fuel 



Btu/ 

Btu/ 




C0 2 

h 2 o 

lb fuel 

lb fuel 

Methane. 

ch 4 

Gas 

Gas 

Liquid 

23,861 


Ethane. 

c 2 h 6 

Gas 

Gas 

Liquid 

22,304 


Butane. 

c 4 h 10 

Gas 

Gas 

Liquid 

21,293 

159 

Pentane. 

c 6 h 12 

Gas 

Gas 

Liquid 

21,072 

158 

Heptane. 

c 7 h 16 

Gas 

Gas 

Liquid 

20,825 

157 

Octane. 

c 8 h 18 

Gas 

Gas 

Liquid 

20,747 

156 

Benzene. 

c 6 h 6 

Gas 

Gas 

Liquid 

18,172 

186 

Decane. 

CioH 22 

Gas 

Gas 

Liquid 

20,638 

154 

Dodecane. 

Ci 2 H 2 6 

Gas 

Gas 

Liquid 

20,564 

154 

Hexadecane. 

c 16 h 34 

Gas 

Gas 

Liquid 

20,472 

154 


* Chiefly from API Research Project 44, National Bureau of Standards. 


In Table 19:1 are shown data regarding the constant-pressure heat of 
combustion (—AH) of various fuels. The heat of combustion is here 
measured at a base temperature of 77°F and is based on phases of fuel 
and of products as indicated in the table. Methods have already been 
discussed by means of which the tabulated heats of combustion may be 
adjusted to different phases of the fuel or of the products. For example, 
let it be required to calculate —AH when butane enters the reaction as a 
liquid and the H 2 0 in the products is a gas. From an earlier computation 
[see Eq. (19:7)], 5 moles, or 90 lb, of H 2 0 are formed in the complete com¬ 
bustion of 1 mole, or 58 lb, of C 4 Hi 0 . From the steam tables, the enthalpy 
of vaporization of H 2 0 at 77°F is 1050-4 Btu/lb. Since heat given up 
owing to condensation of H 2 0 in the products forms a part of -AH as 
tabulated here, we shall deduct ff (1050.4) = 1630 Btu from 21,293, the 
value of -AH from Table 19:1, giving 19,663 Btu per pound of fuel as the 
constant-pressure heat of combustion when the H 2 0 of the products is a 











































474 


BASIC ENGINEERING THERMODYNAMICS 


gas. If the butane had entered the reaction as a liquid, the table shows 
that 159 Btu of the heat generated would have been used in changing it 
to the gaseous phase. The net value of the constant-pressure heat of 
combustion under these new conditions is therefore 19,663 — 159 = 
19,504 Btu per pound of butane. 

The heat of combustion, when based on the assumption that the H 2 0 in 
the products is a liquid, is called the higher heating value; if a gas, the term 
lower healing value is used. With the exception of the combustion of car¬ 
bon, carbon dioxide, and sulfur, where no H 2 0 is formed, all of the values 
of —AH in Table 19:1 are higher heat values. There is of course no dis¬ 
tinction to be made between higher and lower heat value when H 2 0 does 
not appear in the products. 

The heat of combustion at constant pressure may be larger, may be 
smaller, or may be equal to the heat of combustion at constant volume. 
We may write 


— AH = II 


mixture 


- H 


products 


= E m •, 


mixture 


P 

E products + j(V mixture byproducts) 


P 

= — AE + j (V mixture — I products) (19T2) 


Thus it is observed that if the volume of the mixture is greater than the 
volume of the products, when measured at the same pressure and tem¬ 
perature, — AH > —AE. If the two volumes are equal, the two heats of 
combustion are equal, and if the volume of the products exceeds the vol¬ 
ume of the mixture, —AH < —AE. The combustion equation furnishes 
the key as to the relative volumes of mixture and products, as we have 
seen. For example, when the combustion equation is 

H 2 (g) + i0 2 (g) - H 2 0(g) 

the volume of the mixture is greater than the volume of the products, as 
measured at the same pressure and temperature, and —AH > —AE. On 
the other hand, for 

C(s) + i0 2 (g) —> CO(g) 
the opposite is true, while for 

C(s) + 0 2 (g) -> C0 2 (g) 

the volumes are equal, and —AH = —AE. 

This principle may be used to obtain the const ant-volume heat of com¬ 
bustion from the constant-pressure heat of combustion, or vice versa. In 
performing the calculation, it is customary to assume that the gases are 


THERMODYNAMICS OF COMBUSTION 


475 


perfect gases. For example, let it be required to find the constant-volume 
heat ol combustion that corresponds to the following combustion equation: 

C 4 H 10 (g) + 6i0 2 (g) -> 4C0 2 (g) + 5H 2 0(1) 

— AH is given for this reaction in Table 19:1 as 21,293 Btu per pound of 
C 4 Hi 0 . In the combustion of 1 mole, 58 lb, of the fuel, there are 7\ moles 
of the mixture, 9 moles of the products; of the products, 4 moles are in the 
gaseous phase. This represents a reduction of 3-J- moles in volume as a 
result of the reaction. The product PV for 1 mole of a perfect gas is [see 
Art. 9:2 and Eq. (9:2)] 

PV = Pmv = 1545 T 

Thus in Eq. (19:12), 


P 

J 


(V 


mixtu re 


V 


products 


) = 


(PV) mixture - (PV) products 

~~J~ 

(3i) (1545) (460 + 77) 
778 


= 3740 Btu per mole C4H10 
= 64 Btu per pound of C4H10 
Substituting this value in Eq. (19:12), 

Q v = -AE per lb = -AH - 64 = 21,293 - 64 = 21,229 Btu/lb 

Example 19:4. For the coal of Example 19:2, calculate (a) the higher heat value in 
constant-pressure combustion; (6) the lower heat value in constant-pressure com¬ 
bustion; (c) the higher heat value in constant-volume combustion. The base tem¬ 
perature is 77°F. 

Solution: 

(а) The weight of carbon per “mole” of coal is 78 lb. The heat value of this carbon, 
based on Table 19:1, is (78)(14,087) = 1,098,786 Btu per mole of coal. The hydrogen 
in the H 2 0 is, of course, not available for combustion. The rest of the hydrogen in the 
coal, because of its affinity for oxygen, is also assumed to be in combination with the 
oxygen of the fuel to the extent that the amount of oxygen permits. In the formation 
of H 2 0, 8 lb of oxygen unites.with each pound of hydrogen. Therefore, from the 3 lb 
of hydrogen in a mole of this coal, we must deduct of the weight of oxygen it con¬ 
tains, or § = 0.25 lb. (This is often expressed in formulas as H — 0/8.) The 
higher heat value of the remaining 2.75 lb of hydrogen is (2.75) (60,958) = 167,635 
Btu per mole of coal. The single pound of sulfur in 1 mole of the coal will contribute 
4344 Btu to the heating value of each mole. The relative insignificance of this con¬ 
tribution will be noted. Totaling, the higher heat value per mole of coal is 

1,098,786 + 167,635 + 4344 = 1,270,765 Btu per 100 lb = 12,708 Btu per pound 
of coal 

(б) In the combustion of 100 lb of coal, 1.72 moles, or 31 lb, of H 2 0 appears in the 
products (see Example 19:2). The enthalpy of vaporization of this weight of water 
vapor at 77°F is (31)(1050.4) = 32,600 Btu per mole of coal. The lower heat value 
of the coal is therefore 


1,270,765 — 32,600 = 1,238,165 Btu per 100 lb = 12,382 Btu per pound of coal 





476 


BASIC ENGINEERING THERMODYNAMICS 


(c ) Returning to Example 19:2, it will be seen that the number of gaseous moles in 
the reactants is 7.22 + 27.15 = 34.37, while, for-the products, it is 6.5 + 0.03 + 
27.18 = 33.68. (The H 2 0 in the products is not counted as a gas since the higher 
heat value is desired.) The small decrease of 34.37 — 33.68 = 0.69 mole per 100 lb 
of coal is caused by the presence of uncombined hydrogen in the coal. Note that this 
difference would be unchanged if excess air were used to ensure complete combustion, 
for the number of gaseous moles in reactants and products would both be increased 
in the same amount. Then 


J 


(Fmixture Fproducts) 


From Eq. (19:12), 


(PE)mixture ~ (PTOproducts _ (0.69) (1545) (537) , 

J " 778 

= 853 Btu per 100 lb = 8.53 Btu per pound of coal. 


— AE = -AH ~j(V mixture V products) = 12,708 - 8 = 12,700 Btu/lb. 


19:5. The heat of formation is the enthalpy of a compound relative to 
that of the elements of which it is composed; it is stated with reference to 
a selected base temperature. It is thus equivalent to the constant- 
pressure heat of reaction (note that this is not necessarily a “ combustion ”) 
when the reactants consist of the elements and the products exclusively 
of the compound. Thus the heat of formation of C0 2 , when the carbon 
enters the reaction as a solid, is (12) (14,087) = 169,044 Btu/mole at the 
base temperature of 77°F. This is —AH, and if the enthalpy of solid 
carbon and gaseous oxygen at this temperature are arbitrarily assigned a 
value of zero, the enthalpy per mole of C0 2 at the same temperature is 
—169,044 Btu. This relation may be written in the form of an equation 
as 

Hew + H 0 ,( g ) = H c02(g ) ■+■ 169,044 
or 

— 169,044 = H C o 2 ( g ) — H C ( S ) — Ho 2 ( g ) (H in Btu/mole) (19:13) 
Similarly, based on data from Table 19:1, we may write 

122,890 = Hh.,o(d Hn^g) — 4Ho 2 (g) (19:14) 

If the heat of formation of any hydrocarbon is known, Eqs. (19:13) and 
(19:14) may be used to calculate its constant-pressure heat of combustion. 
Conversely, if the constant-pressure heat of combustion of a fuel com¬ 
pound is known, its heat of formation may be computed. Let it be 
required, for example, to determine the heat of formation of butane. The 
reaction relation for butane which is analogous to, and is obtained in the 
same manner as, Eqs. (19:13) and (19:14) is 

1,237,000 = 4H C o 2 ( g ) “b 5H H2 o(p H C4Hl0 (g) — 6-^Ho 2 ( g ) (19:15) 

Equations 19:13 to 19:15 may be arranged as below, multiplying each 




THERMODYNAMICS OF COMBUSTION 477 

term ol an equation by a constant, if necessary, and/or changing signs to 
give 

+ 1,237,000 = — 4H C o 2 ( g ) — 5 Hh 2 o(d + Hc 4 H 10 (g)+ 0 - 2 -Ho 2 ( g ) 

-676,176 = +4H C02( g) -4Ho 2(g ) -4H c(s) 

-614,450 = +5H H2 o(i) -2iH 02( g) -5H fl2( g) 

Adding, we obtain 

— 53,626 = + H C4Hl0( g) — 4H c(b) -5H H2( g) 

This result indicates that the heat of formation of butane is 53,626 
Btu/lb-mole, or 923 Btu/lb, and may be interpreted as meaning that when 
4 moles of solid carbon unite with 5 moles of gaseous hydrogen to form 
1 mole of gaseous butane, the constant-pressure heat of reaction is 53,626 
Btu. The calculation of the heat of formation from the heat of combus¬ 
tion of a fuel compound, as above, or the calculation of the heat of com¬ 
bustion of a compound when its heat of formation is known, is an applica¬ 
tion of the First Law, as a brief study of the foregoing example will show. 

The heat of formation of fuel compounds may be negative as well as 
positive; for example, the heat of formation for acetylene is —3300 
Btu/lb. When negative, the heat of formation indicates that heat is 
given off as the compound is broken down into its elements. 

Example 19:5. Calculate — AH for acetylene. Assume a base temperature of 
77°F and the H 2 0 in the products to be a liquid. 

Solution. One mole of C 2 H 2 contains 24 lb of carbon, 2.016 lb of hydrogen. If 
burned to C0 2 (g) and H 2 0(1), respectively, the constant-pressure heats of combustion 
of these weights of carbon and hydrogen would total 

(24) (14,087) + (2.016) (60,958) = 460,979 Btu 

But the heat of formation of C 2 H 2 is —3300 Btu/lb, or —86,000 Btu/mole. There¬ 
fore 86,000 Btu is absorbed in the formation of 1 mole of C 2 H 2 . According to the 
First Law, this heat must be added to the heat of combustion of the elements C and H 
as calculated above to obtain the heat of combustion of the compound, or 

-Atf C2 H 2 = 460,979 + 86,000 = 546,979 Btu/mole = 21,000 Btu/lb 

19 :6. Dissociation. The direction of the arrow in the basic combustion 
equations of Art. 19:2 [Eqs. (19:1) to (19:5)] allows the reader to infer 
that these reactions are unidirectional. At atmospheric, or even con¬ 
siderably above atmospheric, levels of temperature this is practically true, 
but at high temperatures an increasing tendency toward dissociation 
makes its appearance. Dissociation is the opposite of the combustion 
reaction and can be represented by the same equation but with the arrow 
reversed; at very high temperatures molecules may even dissociate into 
atoms. The First Law will make it clear that, just as the combustion 
reaction was accompanied by the release of energy, energy will be absorbed 
during dissociation. 


478 


BASIC ENGINEERING THERMODYNAMICS 


The temperatures with which the engineer deals do not even approach 
levels at which dissociation is complete, i.e., at which the products of com¬ 
bustion, formed at lower temperatures, are again completely reduced to 
the reactants through the process of dissociation. In the calorimeters 
described in Art. 19:4, essentially complete combustion is ensured by the 
low level of temperature at which the reaction is conducted and by the 
excess of oxygen for combustion that is always provided. Even in the 
common open-flame engineering combustion processes that are carried 
out at considerably higher temperature, as in the furnace of the steam 
power plant, dissociation is usually a negligible factor in reducing the net 
amount of energy release. On the other hand, when combustion is car¬ 
ried out in a confined space under high pressure and when the supply of 
oxygen for combustion is limited, as in the combustion that takes place 
within the cylinder of the internal-combustion engine, the effects of dis¬ 
sociation assume increased engineering importance. 

As the combustion process proceeds, a large part of the energy released 
is temporarily stored in the products, raising their temperature. As a 
result of this increase in temperature, the tendency toward dissociation 
increases, thus slowing down, and finally bringing to a halt, the combus¬ 
tion process before it is completed. When this balance is reached, certain 
amounts of unburned reactants will remain among the products but the 
mixture is said to be in chemical equilibrium, meaning that it has no 
further tendency toward either combustion or dissociation. 

The proportions of fuel that will remain unburned in the products in 
spite of the presence of oxygen in those products depend upon the tem¬ 
perature and pressure, when the mixture is in chemical equilibrium. 1 It 
will also depend upon the amount of oxygen supplied as compared with 
the requirements for complete combustion. These proportions constitute 
the most stable state of the mixture, and the capacity for energy release by 
the fuel as the result of combustion should be based on this most stable 
state. 

In identifying the most stable state, use is made of a parameter called 
the equilibrium constant, which defines the degree of completeness of the 
reaction. This equilibrium constant is computed from a formula based 

1 Fuel and air alone may of course be the sole components of a mixture that is in 
“equilibrium” at atmospheric temperature, or even at levels of temperature con¬ 
siderably above that of the atmosphere. It requires a finite disturbance, such as the 
creation of a high temperature in a very small part of the mixture, to initiate the reac¬ 
tion and to cause the system to assume a state of chemical equilibrium. The non¬ 
chemical equilibrium of the original mixture of fuel and air may be compared to the 
equilibrium of a block held stationary on an inclined plane under the passive resist¬ 
ance that is offered by static friction. Here again, if the necessary disturbance is 
supplied, the system will sweep to a state of stable equilibrium. In both cases, the 
original state is classed as metastable. 


THERMODYNAMICS OF COMBUSTION 


479 


on the partial pressures of the reactants and the products in the mixture 
that results from combustion; for a given reaction it is a function of the 
temperature. To find the degree of dissociation corresponding to a given 
temperature in a given reaction, resort is made to a chart from which the 
value of the constant that corresponds to that temperature and that 
reaction is read. The reaction equation is then written in terms of an 
unknown ratio x which represents the amount of dissociation. For exam¬ 
ple, in the combustion of hydrogen with 100 per cent excess oxygen, this 
equation would be written as 

Hi + O s -> (1 - *)H 2 0 + xH 2 + 0 2 

Based on this equation, the partial pressures of the gaseous products may 
be calculated in terms of x, if these gases are assumed to be perfect gases. 
An expression for the equilibrium constant, written in terms of x, can thus 
be obtained, and, placing this expression equal to the value which has been 
read from the chart, the value of x (the degree of dissociation) may be 
determined; the most stable state corresponding to the given temperature 
can thus be identified. 

The application of this method is attended by complications such as the 
consideration that the final temperature of the products must be con¬ 
sistent with the amount of energy that can be released in changing to the 
most stable state. Also, the fuel may contain both carbon and hydrogen, 
and thus two reactions, instead of one, will be carried out simultaneously; 
this results in two unknowns instead of one and often causes the computa¬ 
tions to be of a trial-and-error nature. A more complete discussion is 
beyond the scope of this book, and the reader is referred to other texts. 1 

In making a thermodynamic study of the processes through which the 
working fluid progresses as it passes through an internal-combustion plant, 
it is essential to locate the most probable state corresponding to the condi¬ 
tions that exist at the beginning and end of each process. If we use the 
Otto-cycle engine as an example, the state at the beginning of the isen- 
tropic compression may be based on the proportions of fuel to air and the 
pressure and temperature at that point, known or assumed; this is not a 
most stable state since the temperature is below that required to initiate 
the combustion reaction. 

Nor is the state at the end of the compression a most stable state, but 
it may be located, if the compression ratio is known or assumed, by apply- 

1 For more detailed discussion of the equilibrium constant, see E. F. Obert, “Ther¬ 
modynamics,” pp. 382-387, McGraw-Hill Book Company, Inc., New York, 1948; 
J. H. Keenan, “Thermodynamics,” pp. 457-465, John Wiley & Sons, Inc., New York, 
1941. 



480 


BASIC ENGINEERING THERMODYNAMICS 


ing the assumption that the system is a simple system for the period of 
the compression and that the compression is isentropic. 

During the constant-volume combustion, the working system sweeps 
to a most stable state which may be located by methods to which reference 
has already been made. As the products begin their isentropic expansion, 
the temperature and pressure decrease; the lower temperature corresponds 
to a lesser degree of dissociation, and there may be some afterburning. 
However, when the temperature falls below about 1600°F in the course of 
this expansion, the reaction rate becomes very slow and the products 
maintain the same ratio of concentration during any further decrease in 
temperature; the system thus may again depart from a most stable state. 
The composition of the products at the end of the adiabatic expansion is 
established, however, and the corresponding state may be located. 

Treating the working system as a simple system from this point until 
the starting point is again reached, the state at the beginning and end of 
the rest of the processes is located and the operation of the simplified 
Otto-cycle engine is analyzed; the qualification “simplified ’’ is made 
since, .although this analysis will show the effects of dissociation and will 
avoid the greater approximations involved in air-standard methods of 
analysis because of assuming the system to be a simple system throughout 
the cycle, it makes no allowance for factors such as friction or radiation. 
Hershey, Eberhardt, and Hottel 1 have constructed combination ES and 
HS charts for various fuel-air ratios which are convenient for a rapid 
solution of problems of this kind. At each level of pressure and tempera¬ 
ture, the internal energy and the enthalpy of the most stable state of the 
products may be read from these charts. 

19:7. The flame temperature is the highest temperature reached by 
the gases as the combustion is carried to the most stable state of the prod¬ 
ucts. In the solution of problems in which dissociation is an important 
factor, much time may be saved if the engineer can make an accurate 
estimate of the flame temperature that will result. 

The easiest, but the most inaccurate, assumption that can be made as 
the basis of an estimate of flame temperature is that combustion is com¬ 
plete, that the specific heat of the products is constant, and that the com¬ 
bustion process is adiabatic. If the combustion takes place at constant 
volume, the work, as well as the heat flow, is zero and the internal energy 
of the products must equal the internal energy of the mixture. But, at 
the same temperature, these internal energies differ by an amount which 
we have called the constant-volume- heat of combustion and designated 
as Q v . Q„ must therefore equal A U for the products as their temperature 

1 R. L. Hershey, J. E. Eberhardt, and H. C. Hottel, “Thermodynamic Properties of 
the Working Fluid in Internal-combustion Engines,” SAE Journal , October, 1936, 
pp. 409-424. 


THERMODYNAMICS OF COMBUSTION 


481 


increases to the flame temperature from the temperature at which the 
reaction began, or 

At products = M p C Vp (T 2 T l) 

and 


T 2 



Q v 

AI pC Vp 


(19:16) 


in which T 2 = flame temperature 

Ti = temperature at which reaction began 
M p = mass of products per unit mass of fuel 
c Vp = specific heat at constant volume of the products 
If the combustion takes place at constant pressure, as in open-flame 
steady-flow combustion, there is again no work, other than flow work, and, 
from Eq. (3:5), we see that Q p must equal AH for the products, or 


T 2 


T ! 



NI pC Pp 


(19:17) 


in which the meaning of the terms is the same as in Eq. (19:16) except 
that Q p , the constant-pressure heat of combustion, has replaced Q v and 
c Pp , the specific heat at constant pressure of the products, takes the place 
of c Vp . Q p differs only in relatively small amount from Q v ; it may be 
larger, smaller, or have the same value. On the other hand, c p for the 
gases in the products is considerably larger than c v for those products. 
It becomes evident that the flame temperature that results from a con¬ 
stant-pressure combustion will be considerably below that resulting when 
the combustion takes place at constant volume. 

In the foregoing analyses, the specific heats have been held constant 
for simplicity of treatment. If their variation is to be given effect, 

A U = Mp j Ti c Vp dT and AH = M p J T * c Pp dT. Since A U = U 2p Toducts 

— TJ iproducts and AH = H 2p ro duct s — H i product8 , the tables of internal energy 
and enthalpy which are included in the gas tables may be used to simplify 
the estimation of flame temperature. 

The flame temperatures as calculated by the method described above 
are highly theoretical and represent a maximum that could conceivably 
be attained only on the basis of the assumptions which have been made. 
The engineer makes deductions from this theoretical flame temperature 
to allow for various factors in arriving at his estimate of a probable actual 
flame temperature. Some of these factors are listed below: 

1. As the flame temperature increases, the effect of dissociation becomes 
increasingly important. This effect is to reduce the net, or effective, heat 
of combustion per unit weight of the products and thus to reduce the 
flame temperature. As has been pointed out above, flame temperatures 




482 


BASIC ENGINEERING THERMODYNAMICS 


tend to be higher for constant-volume combustion, and the effect of dis¬ 
sociation is correspondingly greater. 

2. The temperature of the products is above the temperature of the 
surroundings, and a leakage of heat to the surroundings results. The 
real combustion is not instantaneous and therefore not adiabatic, as was 
assumed above. Again the effect is to lower the flame temperature. 

3. Because the combustion is not instantaneous and because it takes 
place in the internal-combustion engine behind a rapidly moving piston, 
neither constant-volume nor constant-pressure combustion is actually 
attained in these engines in practice. If the volume increases even 
slightly during what is nominally a constant-volume combustion, the 
flame temperature is correspondingly reduced; if, during a “constant- 
pressure” combustion, the pressure drops off before the reaction is com¬ 
plete, the flame temperature is lowered. 

19 :8. Power from Combustion. The engineer makes use of combus¬ 
tion in the manufacture of power to create a source from which heat may 
be withdrawn at high temperature to supply the heat engine. It has been 
pointed out in the introduction to the present chapter that only a portion 
of the heat of combustion may be delivered at temperatures at which this 
heat may be efficiently used by the heat engine. In preceding chapters it 
has been emphasized that, even of the heat which reaches the heat engine, 
only a portion may conceivably be converted into work. The engineer 
thus realizes a maximum of only about 30 per cent of the heat of combus¬ 
tion in the form of work. 

By returning to Art. 6:10, it may be shown that a far larger work 
delivery is at least thermodynamically conceivable. In that article it 
was brought out that the maximum useful work that could be delivered 
by a system, as it changed between two states in each of which it was in 
pressure and temperature equilibrium with the atmosphere, was equal to 
the decrease of the zeta, or Gibbs free-energy, function; to realize this 
maximum useful work, it was required that the change of state be accom¬ 
plished reversibly and that heat be exchanged only with the atmosphere. 

Essentially, the fuel and air enter the heat engine at the pressure and 
temperature of the atmosphere; in the limit, the products of combustion 
reach the same pressure and temperature as they are returned to the 
atmosphere. If the necessary conditions are observed, the combustion 
process may therefore conceivably account for a maximum amount of 
useful work, Which is defined by the equation 

= Z X -Z 2 = (H x - T 0 Si) - {H 2 - T 0 S 2 ) =H l -H 2 - T 0 (S l - S 2 ) 

(19:18) 

in which the subscript 1 refers to the reactants, 2 to the products, and T 0 
is the absolute temperature of the atmosphere. The difference Hi — II 2 


THERMODYNAMICS OF COMBUSTION 


483 




is a maximum if sufficient oxygen is provided for complete combustion 
and if all of the carbon in the fuel appears in the products as carbon 
dioxide, all of the hydrogen as II 2 0; this difference is the —AH of Eq. 
(19:11), the heat of combustion at constant pressure. The difference 
Zi — Z 2 , measuring the maximum useful work, will therefore approach 
the heat of combustion at constant pressure as the entropy of the products 
approaches that of the original mixture. 

\\ hen the engineer uses combustion merely as a source of heat, he with¬ 
draws heat in large amounts from the products at high temperatures and 
thus fails to meet the conditions under which Eq. (19:18) is applicable. 
In order to meet those conditions, all operations must be reversible, and 
heat exchange can take place only with the atmosphere; let us see how 
these requirements might be satisfied. Consider, for example, a mixture 
of solid carbon with the amount of oxygen necessary for its complete com¬ 
bustion at the temperature and pressure of the atmosphere. Now let 
this mixture be isentropically compressed to a temperature high enough 
so that dissociation will be complete. In the course of this compression, 
the system will pass through the intermediate temperatures, and if com¬ 
bustion is to be prevented during this operation, some sort of inhibitor, or 
anticatalyst, must be provided. However, once the temperature at which 
dissociation is complete is reached, the inhibitor can be removed and there 
will be no unbalanced tendency for the carbon and oxygen to combine. 

The inhibitor having been removed, the mixture is allowed to expand 
slowly and isentropically. As the temperature drops slightly in the first 
stage of this expansion, a new most stable state is reached which permits 
the combustion of a limited portion of the carbon. As the expansion con¬ 
tinues to lower and lower and lower temperatures, more and more C0 2 is 
formed until, by the time the temperature of the atmosphere has again 
been reached, combustion is essentially complete, with no unburned car¬ 
bon left. The ordinary process of combustion is a sweeping change with 
mounting temperatures which, once started, cannot be reversed, but the 
combustion described here may be conceived as reversible, since, by 
recompression, the products can be returned to the form of the reactants 
over an identical, but reversed, path. 

During the original isentropic compression with an inhibitor present, 
the work of compression was E 0mixture - E' miiture , where E 0mixture is the 
stored energy of the system as a mixture of the reactants at atmospheric 
temperature and EA xture is the stored energy of the same mixture of reac¬ 
tants, but at the higher temperature of complete dissociation. For the 
succeeding isentropic expansion during which combustion takes place, 
the positive work is E' miIture - E 0producta , the net work of the double process 
thus being E 0mixture - E 0products , or -A E 0 . In the reaction that has been 
used as an example, the volume of the products and reactants are the 


484 


BASIC ENGINEERING THERMODYNAMICS 


$ 


same, and, when a return has been made to the temperature of the atmos¬ 
phere, the pressure of the atmosphere has also been reached, the products 
may be directly rejected to the atmosphere, and a fresh charge accepted. 
Also, — A B 0 = —A H 0 , and, since both processes have taken place without 
change of entropy, W Um&x = — AH. 

For other reactions, the pressure of the system as it reaches the tem¬ 
perature of the atmosphere in its expansion may be above or below the 
pressure of the atmosphere, according to whether the volume of the prod¬ 
ucts is greater or less than the volume of the reactants at the same pres¬ 
sure and temperature. If the pressure is higher than that of the atmos¬ 
phere, an isothermal expansion to atmospheric pressure may take place 
while heat flows reversibly from the atmosphere; this will be accompanied 
by an increase of system entropy so that $ 2 > Si and the useful work 
will, as based on Eq. (19:18), slightly exceed H i — H 2 (= —AH). When 
the pressure is less than atmospheric pressure at the end of the expansion 
to atmospheric temperature, an isothermal compression, accompanied by 
reversible heat flow to the atmosphere and a decrease in system entropy, 
may be used to restore pressure equilibrium; here the maximum useful 
work will be slightly less than —AH. 

The engineer has an example of a similar method of energy conversion 
before him in the ordinary lead storage battery. The inhibitor which 
cancels the tendency toward chemical reaction consists in this device of a 
difference in electrical potential which is maintained between the reactants. 
When current is drawn from the battery, this difference is reduced and 
the necessary reaction takes place to again restore a condition of equilib¬ 
rium. An important difference between this reaction and its combustion 
counterpart as described earlier is that the energy transfers in the battery 
take place at constant temperature, the temperature of the atmosphere. 

If an electric cell to control the reaction between carbon and oxygen 
could be devised, the work obtained from a pound of coal would be greatly 
increased. For reasons such as his failure to discover a suitable inhibitor 
and his inability to attain the temperatures equivalent to complete dis¬ 
sociation in the combustion of any important fuel, the engineer has so far 
been unable to realize the benefits that are indicated as theoretically 
possible. 

Problems 

1. Write relations analogous to Eqs. (19:1a) to (19:ld) but based upon combustion 
equations (19:2) to (19:5), assuming (a) that the oxygen is supplied as pure oxygen 
in exactly the right proportion for the combustion and (6) that it is supplied as a part 
of the ideal amount of air. 

2. A gaseous hydrocarbon series has the formula C„H n+2 , where n is an integer. 
Write the balanced combustion equation for this series in terms of n, assuming (a) that 
pure oxygen is supplied in the minimum amount for complete combustion, (6) that 
combustion is complete with the theoretical air, and (c) that it is complete when 


THERMODYNAMICS OF COMBUSTION 


485 


10 per cent excess air is supplied. ( d ) Assume that 90 per cent of the theoretical air 
is supplied and that no free oxygen, carbon, or hydrogen appears in the products. 

3. Work Example 19:2, but assume a coal with the following analysis: C, 65; H, 4; 
O, 8; S, 1; N, 1; H 2 0, 12; ash, 9. 

4. In burning octane, C 8 Hi 8 , an Orsat analysis of the products shows the following 
percentages: C0 2 , 11; 0 2 , 2.5; CO, 1. What is the air-fuel ratio? The percentage of 
excess air? 

5. In burning the coal of Prob. 3, an Orsat analysis of the products shows the 
following percentages: C0 2 , 11; 0 2 , 8; CO, 1. Assuming the cinder in the ashpit to 
be 40 per cent unburned carbon, write the combustion equation, and determine the 
percentage of excess air. 

6. Calculate the percentages that would be shown by the Orsat analysis of the 
products when C 8 Hi 8 is burned with 10 per cent excess air, if the percentage of CO in 
the products is 5 per cent of the percentage of C0 2 . 

7. Calculate the constant-pressure heat of combustion of octane when (a) the fuel 
enters the reaction as a liquid and the water vapor leaves as a liquid, (b) when both 
are gases, and (c) when the fuel is liquid and the water a gas. 

8. Based on the data of Table 19:1 and assuming that the gases formed are perfect 
gases, calculate the constant-volume heat of reaction of octane(l) at 77°F when the 
water vapor formed is entirely condensed. 

9. Based on a temperature of 77°F and assuming that the water in the products 
is in the form of a gas, the constant-volume heat of combustion of a certain oil is 
17,900 Btu/lb. If the composition of the oil may be expressed as C, t H 2n , when n is an 
integer, calculate the constant-pressure heat of combustion at the same temperature. 
Assume all gases to be perfect gases, the oil to have negligible volume. 

10. Calculate the constant-volume heat of combustion when (a) gaseous octane 
enters the reaction and the water in the products is a liquid and (5) when the octane 
enters as a gas but the water leaves the reaction as a gas. The base temperature is 
77°F. 

11. Calculate the lower heating value of octane at 77°F. 

12. Work Example 19:4, substituting the coal of Prob. 3. 

13. Find the heat of formation of octane. 

14. The heat of formation of ethylene (C 2 H 4 ) is —174 Btu/lb. Find Q p , assuming 
a base temperature of 77°F and that the water in the products is a liquid. 

15. The heat of formation of propane (C 3 H 8 ) is 1440 Btu/lb. Calculate Q p , assum¬ 
ing the standard base temperature and that the water in the products is a liquid. 

16. Assuming that combustion is adiabatic and complete, that the specific heat of 
the products is constant and may be based on Table 9:1, that the water in the products 
is in the form of a vapor, and that the combustion starts at 77°F, calculate the theoret¬ 
ical flame temperature when gaseous octane is burned at constant pressure (a) with 
the theoretical air and ( b ) with 20 per cent excess air. 

Symbols 

c p specific heat at constant pressure 
c v specific heat at constant volume 
E stored energy of a system 
(g) gaseous 
h specific enthalpy 
H enthalpy of a system 
J proportionality factor 
(1) liquid 


BASIC ENGINEERING THERMODYNAMICS 


486 


m molecular weight; weight of 1 mole 
M mass of a system 
P pressure 
Q heat flow 

Q p constant-pressure heat of combustion (= —AH) 
Q v constant-volume heat of combustion (= —A U) 
(s) solid 

T absolute temperature 
T 0 absolute temperature of the atmosphere 
U internal energy of a system 
v specific volume 
V volume of a system 
W work 

x amount of dissociation, a decimal fraction 

Subscripts 

C of carbon 
C 4 H 10 of butane 
C0 2 of carbon dioxide 
H 2 of hydrogen 
H 2 0 of water 
0 2 of oxygen 
p constant pressure 
T constant temperature 
v constant volume 


CHAPTER 20 


THE TRANSMISSION OF HEAT 

20:1. Introduction. Heat has been defined in Chap. 1 as that form of 
energy which moves from the body at higher temperature to that at lower 
temperature by reason solely of a temperature difference that exists 
between them. In intervening chapters we have observed that such 
“heat transfers” are often necessary in carrying out specific thermody¬ 
namic processes. Only casual reference has been made, however, to the 
apparatus necessary to effect these heat exchanges and none at all to 
the basic principles which underly the design of a heat exchanger. If the 
broader definition of the meaning of the word thermodynamics as the 
study of changes in the form and the location of energy is accepted, at 
least some attention should now be given to these principles. 

Heat exchangers are a very important part of the engineering plant. 
Their first cost and the cost of their operation and maintenance are of 
great importance to the designer in developing a plan for an efficient 
installation. Thermodynamically, as we have seen, it is desirable to 
reduce the irreversibility that always accompanies heat transmission in 
practice by reducing the temperature interval between the bodies that 
exchange heat. The cost of heat exchangers increases rapidly with this 
reduction in temperature differential, and the designer must make a 
decision as to the economic limit to which compliance with thermody¬ 
namic principles may go. 

Heat exchange is sometimes accomplished, as in jet condensers and open 
feedwater heaters, by the direct mixing of warm and cold fluids. Where 
this method of removing heat from a hot or adding it to a cold fluid can 
be used, it is most efficient from the standpoint of the size and cost of the 
apparatus required but, on the basis of the Second Law, also is often 
accompanied by the maximum thermodynamic irreversibilities. In a 
more commonly used form of the heat exchanger, the heat is transferred 
between fluids which are separated by a solid wall, such as the wall of a 
boiler tube; the heat transferred between the fluids cannot exceed the 
capacity of the wall for heat transfer, and this capacity is the basis of 
design for this type of heat exchanger. Examples are found in the air 
preheater, the surface condenser, and the economizer. This class of heat 
exchanger, in which the fluids are kept separate, is often the only type 
that is feasible from the practical engineering standpoint when the fluids 

487 


488 


BASIC ENGINEERING THERMODYNAMICS 


differ in their chemical composition and at least one is to be recirculated 
through the plant cycle. 

Provision for heat exchange may be planned and deliberate and may 
take place under carefully controlled conditions, as in the boiler and con¬ 
denser of the power plant. Or the heat exchange may be variable and 
constitute the load on the plant, as in the design of a plant for heating a 
building. Here the building acts as a heat exchanger, the heat losses 
varying in magnitude with weather conditions; the plant must be capable 
of replacing these losses under the most extreme conditions. A similar 
example is found in the refrigerating plant, except that the natural flow 
of heat takes place in the opposite direction in this case; refrigeration 
plants must be designed with a capacity sufficient to remove these heat 
gains. 

The subject of heat transfer has developed in recent years to the point 
where space requirements forbid more than a review of basic principles in 
a text of this character. When the transfer of heat is accomplished by 
the irreversible mixing of fluids, the engineering problem is usually of the 
simplest character, the properties of the mixture being computed by 
equating the energy lost by one fluid to the energy gain of the other. For 
that reason our attention will be centered on the principles which govern 
when heat is exchanged between fluids that occupy separate compart¬ 
ments or flow through separate channels; in that case design should be 
based on a proper balance between the theoretical thermodynamic and 
the economic factors that enter the specific problem. 

The classic method of approaching the study of heat transfer through 
intervening materials or spaces classifies the transfer of heat as being by 
conduction , by radiation , or by convection. The meanings of these terms 
are separately outlined below but, in the form of heat exchanger employed 
in the engineering plant, the transfer of heat takes place typically through 
a combination of these methods. 

Conduction. Heat transfer between a hot and a cold body by conduc¬ 
tion requires that the space between them must be occupied by some 
material substance such as a metal wall; but conduction may take place 
through fluids, either gases or liquids, as well as through solids. The 
classic example of conduction, to which the reader has undoubtedly been 
introduced in elementary texts on physics, is the flow of heat along a 
metal rod, one end of which is placed in a flame, the other held in the 
hand. As the end which is placed in the fire receives heat, the rate of 
random movement of the molecules at that end of the rod is increased; as 
the result of collisions with molecules farther along the rod, they impart 
increased momentum to these adjacent molecules, and the temperature 
gradually increases along the rod until, conceivably, it may become too 
hot to hold. Neglecting the escape of heat into the surrounding air from 


THE TRANSMISSION OF HEAT 


489 


sections of the rod between its two ends (as if these sections had been 
covered by some sort of shield which prevented the escape of heat at an 
angle to the rod axis), the time rate at which the temperature will increase 
at the cold end will depend upon the length of the rod, the differential of 
temperature between its two ends, and the capacity of the material of 
which the rod is composed for transferring molecular momentum along 
its length. This rate of increase of temperature is a measure of the rate 
of heat transfer along the rod. Eventually it is conceivable that a bal¬ 
anced condition will be attained, after which the temperature at neither 
end of the rod, nor at any intermediate section, will change. This is a 
. steady state , and thereafter the flow of heat will continue at a rate which is 
independent of time. The design of heat exchangers is normally based 
on the assumption that a steady state is attained, and our discussions to 
follow will assume this condition whether that discussion is related to a 
transfer of heat by conduction, by radiation, by convection, or by a com¬ 
bination of methods. 

The rod temperatures that are reached as the steady state is attained 
will depend upon a number of factors, including not only the capacity of 
the rod for the transfer of heat by conduction but also the capacity of its 
hot end for receiving heat from the fire and the ability of the hand, or 
other external body, at the cold end to retransfer the heat which it 
receives without further increase of temperature at that end. Based upon 
the First Law, the rate at which heat escapes at the cold end and the rate 
at which heat enters at the hot end must both be equal to the time rate of 
heat transfer by conduction along the rod. This equality constitutes a 
basic relation in heat-exchanger design. 

Radiation. Heat may be transferred between two separated bodies 
even when no material substance fills the space between them; this is 
called heat transfer by radiation. All bodies radiate heat in the form of 
an electromagnetic wave motion similar to the waves that transmit light. 

Suppose a body to be placed in a vacuum and isolated from all other 
bodies; the sun of our solar system is an adequate though not entirely 
accurate example, unless we imagine it as the sole occupant of limitless 
space. Radiation will take place from the surface of this body in the 
form of heat and will travel through the surrounding spaces in straight 
lines, much as a light wave might travel. Now let a second body be 
introduced into this surrounding space; an obvious example is our own 
planet, the earth. The earth will also be the source of heat waves into 
the space that surrounds it. A part of the heat waves from the sun will 
be intercepted by the earth and, striking its surface, may be reflected or 
absorbed or may pass through the earth into the space beyond it; the 
latter effect is negligible in the case of the earth. At the same time, the 
earth’s heat waves are being intercepted at the sun’s surface, and an 


490 


BASIC ENGINEERING THERMODYNAMICS 


exchange of energy, though by no means an equal exchange, thus occurs. 
If we use the earth as our “thermodynamic system,” the net amount of 
heat which it receives by radiation is the difference between the energy 
carried along the small fraction of the sun’s rays which it intercepts and 
the energy it sends out in similar rays from all of its surface. Because the 
intensity of heat radiation has been shown to be proportional to the fourth 
power of the absolute temperature of the surface from which the rays 
emanate, the earth receives about as much heat by radiation from that 
part of the sun’s rays that it intercepts as it sends out into surrounding 
space from its entire surface; this balance varies slightly with the season 
of the year over limited portions of the earth’s surface. The net heat 
flow by radiation is thus almost zero with respect to the system that has 
been selected. 

Returning to the example of the rod that was used in discussing heat 
transfer by conduction, it is probable that no small part of the heat that 
flowed into the hot end of the rod was radiated heat. The amount of 
radiant-heat transfer between two bodies becomes important when the 
temperature of one of the bodies is high or both temperatures are high 
but a significant difference in temperature exists between them. In the 
example, the temperature of the flame is high enough to ensure fairly high 
intensity of its heat-wave emanations. Of course only a small fraction of 
these rays are directly intercepted by the rod end and transmitted as con¬ 
ducted heat along the rod. The rod would have received heat by radia¬ 
tion even if it had been removed from direct contact with the flame, but 
the number of rays it intercepted would have been fewer and the heat 
transfer proportionately less in this case. Note also that these radiant 
waves would in this case have passed through a material substance, the 
air surrounding the rod end. A void is not necessary for the transfer of 
heat by radiation; some substances, especially the more transparent 
solids and gases, transmit, rather than absorb or reflect, by far the larger 
portion of heat rays that they intercept. Thus the temperature of the 
air, even a very short distance in front of a brick wall on which the sun’s 
rays fall directly, may be considerably below the temperature of the wall 
surface. 

Convection. The passage of heat purely by conduction assumes that 
the two bodies between which heat is exchanged and the intervening wall 
of material through which that heat must pass have no relative movement 
with respect to each other. If either body, or the wall between them, 
consists of a fluid, it is probable that the heat flow will produce differences 
in temperature, and thus in density, of various sections of the fluid and 
that relative movement of the layers of fluid will result. When fluid 
movement occurs in this manner it is called natural convection. Often, 
as, when the fluid is pumped or blown through the passages of a heat 


491 


THE TRANSMISSION OF HEAT 

I 

exchanger, its movement is directly encouraged and the term forced con¬ 
vection is used. In either case, the rate of heat flow by conduction may 
be considerably affected, and the heat transfer is classified as being by 
convection. In engineering forms of the heat exchanger, the transfer of 
heat is typically between fluids and convective effects are of great 
importance. 

As an example of the effect of convection on heat transfer, consider a 
vertical building wall, as illustrated in Fig. 20:1. The air outside the 
building, at the lower temperature t 0 , and the air inside, at U, will first be 



Fig. 20:1. Temperature gradient through a vertical homogeneous building wall. 

assumed to be stationary with respect to the wall, with all convective 
effects eliminated. The temperature of the air outside and inside the 
wall are both relatively low, and heat flow by radiation is negligible; the 
sole method of heat transfer is therefore by conduction in the assumed 
case, which is illustrated in part a of the figure. For simplicity, it is 
assumed that the flow of heat takes place from a vertical layer of air a 
distance Xi from the inside surface of the wall and that the temperature of 
this layer is U. A temperature difference of U — ti between this layer and 
the layer of air in immediate contact with the inner wall surface furnishes 
the incentive for the flow of heat toward the wall. The temperature 
gradient is this temperature difference divided by the distance Xi, or 
(ti — ti)/xi. On the outer side of the wall a temperature gradient 
(t 2 — to)/x 0 accounts for the flow of heat from the wall into the outside 
air. The slopes of the lines Ut\ and Ut 0 are graphically representa¬ 
tive of the temperature gradients, and these slopes have been shown as 
equal under the conditions assumed in (a) of Fig. 20:1 since the heat flow 



















492 BASIC ENGINEERING THERMODYNAMICS 

% 

to and from equal inside and outside wall areas must be the same. The 
heat flow through the wall is also identical per unit of time with the heat 
flow into its inner and away from its outer surface, and this heat flow is 
secured by means of the temperature gradient (fa — t 2 )/x through the 
wall. In this connection, it will be noted from the figure that (1) the skin 
temperatures fa and fa of the wall have been shown as equal to the tem¬ 
peratures of the air layers in immediate contact with the respective sur¬ 
faces and (2) the temperature gradient through the solid wall has been 
shown as much smaller than that which it is necessary to maintain through 
the air layers in order to produce the same flow of heat by conduction; 
this means that the solid material of the wall has been supposed to be a 
better agency for the transfer of heat by conduction than the air on either 
side and is a natural assumption since the more densely crowded molecules 
of the wall would appear to be capable of imparting their momentum 
more rapidly to adjacent molecules, collisions being more frequent. 

Now let us introduce the effects of convection, as in Fig. 20:16. The 
difference of temperature between the layers of air at fa and fa will bring 
about a difference in their densities, and the layer of air in contact with 
the wall’s inner surface will flow downward, the layer at fa upward. Thus 
a circulation will be produced, and the resultant mixing will have a 
tendency to reduce the temperature difference fa — fa. The assumption 
made in case a would lead us to expect a maximum air temperature at 
some point near the middle of a room exposed on all four sides, but the 
circulation that results from convective effects causes a plateau of tem¬ 
perature to be reached at a short distance from the inner wall surface, as 
is indicated by the leveling off of the temperature-gradient curve at a 
point very close to the wall. The temperature-gradient curve is no longer 
of constant slope but has its greatest slope close to the wall. A similar 
effect is meanwhile being produced by convective effects at the outside 
wall surface; in this case, the skin temperature of the outside wall surface 
is lowered. Note in comparing the two figures that the lowering of fa has 
been greater than the temperature range through which fa was raised. 
This is based on the assumption that wind in the outside air will act to 
increase the convective effects, the air-layer thickness necessary to attain 
the plateau level of outside temperature being less than the thickness of 
the similar layer at the inside wall. The effect at the inside wall may be 
ascribed to natural convection, that on the outside surface may be com¬ 
pared to the forced convection used in many engineering forms of the heat 
exchanger. 

One of the effects of convection has been to increase the temperature 
gradient through the wall; for the temperature difference fa — fa is now 
greater than when no allowance was made for convection. The flow of 
heat through a unit area of wall surface in unit time is thus increased in 


THE TRANSMISSION OF HEAT 


493 


proportion to the increase of this temperature differential. But the heat 
entering and leaving the wall must also have increased if steady-state 
conditions are assumed. Accordingly, as has been shown in Fig. 20:16, 
the temperature gradient in the air layers immediately adjacent to inner 
and outer wall surfaces has been increased above the comparable slopes 
as shown in part a of that figure to account for this greater rate of heat 
flow. 

The engineering heat exchanger typically transfers heat between fluids, 
and the effects of convection, natural or forced, are of corresponding 
importance in design. The problem of calculating convective heat trans¬ 
fer is much more varied, complicated, and difficult than for either con¬ 
ducted or radiant heat. For the purpose of illustrating the effects of 
convection, we have here made use of a very simple example. It will 
occur to the reader that a number of factors will enter the problem and 
exert their influence in controlling the final result. Among these are, 
with respect to the solid wall, its length in the direction of convective flow 
and its inclination to the vertical. With respect to the fluid, some of the 
important factors are the density, the heat capacity (specific heat), the 
effect of temperature differences on the density (the coefficient of cubical 
expansion), the viscosity, and its capacity for conveying heat by conduc¬ 
tion. Also to be considered are the temperature difference between fluid 
and wall and, in the case of forced convection, the velocity of flow. 

In conclusion, it should be emphasized that convection is not a basic 
method of heat transfer, as are conduction and radiation. But the effects 
of convection are to change, often in radical degree, the rate of heat trans¬ 
fer by conduction. This effect is accomplished by increasing the tem¬ 
perature gradient through both the wall and the immediately adjacent 
fluid layers. It has been found convenient, however, to handle its dis¬ 
cussion separately from the discussion of heat transfer by conduction and 
radiation. 

20:2. Heat Transfer by Conduction. The basic equation for heat 
transfer by conduction follows: 

Qcond = IcA ^ (20:1) 

in which Q cond is the heat transferred per hour across a wall of area A nor¬ 
mal to the direction of heat flow, through a material which has a thermal 
conductivity A;,* when the temperature gradient is dt/dx. No difficulty 
will be encountered on account of algebraic sign in the use of this equation 
if it is remembered that the direction of heat flow must always be toward 
the lower temperature. The units in which each of the terms of Eq. 

* This symbol should not be confused with the ratio c p /c v , also represented by the 
same letter. 


494 


BASIC ENGINEERING THERMODYNAMICS 


(20:1) are expressed must, of course, be mutually consistent. It is the 
usual practice, in the English system, to express Q cond in Btu per hour, A 
in square feet, t in degrees Fahrenheit, and x in inches. The unit of 
measurement of the thermal conductivity k would then be the number of 
Btu of heat conducted through a homogeneous wall of the particular 
material 1 in. in thickness in 1 hr when the differential of temperature 
across the wall was 1°F. A unit of thermal conductivity for which the 
basic wall thickness is 1 ft is also important and is one-twelfth the size of 
k as described above, since the resistance offered to the flow of heat is 
twelve times as great for twelve times the wall thickness. When the con¬ 
ductivity is expressed in terms of this unit, the dimension of x changes to 
feet, and the temperature gradient increases in the same proportion so 
that the same heat flow is calculated. The basic dimension of thermal 
conductivity is always the same and may be derived from Eq. (20:1) as 
follows: 



Q dx 

[ML 2 /0 3 ][L] 

ML 

A dt 

[LW) 

d*T_ 


( 20 : 2 ) 


It will be noted that it has here been convenient to add a fourth basic 
dimension T, representing temperature, to the three suggested in Chap. 
17. 

20:3. Thermal Conductivity. Not only solids but also liquids and 

gases are capable of transferring heat by conduction. The conductivities 
of homogeneous solid materials are relatively high, tending to increase 
with their density and their elasticity and, for a specific material, with the 
temperature and the absorption of moisture. Good insulation against 
the conduction of heat is provided by solids which are porous, fibrous, or 
cellular. Values of k may be obtained from tables similar to Table 20:1. 
The variation of thermal conductivity of solids with temperature is not 
ordinarily shown in these tables since their resistance to the flow of heat is 
usually only a small part of the total resistance encountered by that flow 
in the engineering heat exchanger. Values of k vary widely for solids, 
from a minimum of about 0.3 (Btu) (in.)/(ft 2 )(°F)(hr) for the best 
insulators to a maximum of around 3000 for the best conductor of heat. 

The thermal conductivity of liquids and gases is not as accurately 
measurable as that of solids because of the difficulty of eliminating con¬ 
vection currents in fluids. A larger variation in experimental results and 
in tabulated values of k may therefore be expected. The variation of 
conductivity with temperature is of greater engineering importance for 
fluids, especially for gases, and a linear relation between k and t is employed; 
thus 


k — fcofl + Ci(t — * 0 )] 


(20:3) 









THE TRANSMISSION OF HEAT 


495 


Table 20.1. Thermal Conductivities of Some Solids, Liquids, and Gases* 


Material 


Solids 

Aluminum. 

Aluminum, piston alloy. . . . 
Asbestos, corrugated board 

Asphalt. 

Bearing metal, white. 

Brass. 

Brick, building. 

Fire-clay. 

Cast iron. 

Celotex. 

Concrete, stone. 

Copper. 

Corkboard. 

Felt, wool. 

Glass, window. 

Glass wool. 

Gold. 

Gypsum plaster. 

Lead. 

Limestone. 

Magnesia, 85%. 

Mineral wool. 

Scale, boiler. 

Silver. 

Soil, dry. 

Wet. . 

Steel. 

Wallboard, fiber. 

Plaster. 

Wood, balsa. 

Fir and pine. 

Oak and maple. 

Liquids 

Ammonia. 

Benzene. 

Brine (NaCl), 25%. 

Kerosene. 

Mercury. 

Petroleum oil, sp gr = 0.88. 

Sulfur dioxide. 

Water, pure. 

Sea. 

Gases 

Air. 

Ammonia. 

Carbon dioxide. 

Carbon monoxide. 

Hydrogen. 

Nitrogen. 

Oxygen. 

Steam. 


1 

o 

0 

k 0 , (Btu)(in.)/ 

(ft 2 ) (°F) (hr) 

Ci 

32 

1000 

For solids, 
use 0 

400 

1200 


86 

0.84 


70 

4.8 


68 

164 


32 

590 


68 

5 


900 

7.2 


68 

350 


70 

0.33 


• . . 

6.5 


32 

2100 


86 

0.3 


86 

0.36 


68 

6 


... 

0.27 


64 

2028 


• • • 

3.3 


64 

2400 


75 

6.5 


212 

0.47 


86 

0.27 


150 

10-20 


32 

3200 


95 

1.0 


105 

4.0 


200 

300 


70 

0.34 


86 

0.48 


86 

0.36 


• • • 

0.8 



1.15 


68 

4.03 

+0.0018 

32 

1.02 

-0.0028 

-4 

2.76 

+0.0006 

68 

1.04 

-0.0009 

32 

58 

0 

0 

0.94 

-0.0003 

68 

2.3 

+0.0003 

68 

4.1 

+0.001 

68 

5.0 


32 

0.163 

+0.00165 

32 

0.149 

+0.00028 

32 

0.097 

+0.0021 

32 

0.155 

+0.0017 

32 

1.13 

+0.0015 

32 

0.168 


32 

0.170 


32 

0.117 

+0.00217 


* Adapted from various sources, including International Critical Tables; American Society of Refrig¬ 
erating Engineers, “Heat Transmission of Building Materials”; W. J. King, Mech. Eng., April, 1932; 
W. H. McAdams, “Heat Transmission,” 2d ed., McGraw-Hill Book Company, Inc., New York, 1942; 
American Society of Heating and Ventilating Engineers, “Guide.” 



































































496 


BASIC ENGINEERING THERMODYNAMICS 


in which k is the thermal conductivity at temperature t, k 0 at some base 
level of temperature to, and C\ is a constant which defines the relative rate 
of change of conductivity with temperature; C 1 may be positive or nega¬ 
tive. Values of fc 0 , the corresponding temperature to, and of the constant 
Ci are shown for some liquids and gases in Table 20:1. 

20:4. Typical Problems in Heat Transfer by Conduction. The funda¬ 
mental problem in the transfer of heat by conduction is the calculation 
of the heat transferred through a homogeneous plane wall such as that 
illustrated in Fig. 20:1. A variation of this problem is found when the 
wall, though bounded by parallel plane surfaces, consists of laminations 
of materials having different conductivities. Another problem presents 
itself when the wall is not plane but is curved so that the areas through 
which the heat enters the wall and leaves it differ; the most common 
engineering example of this problem is the calculation of the heat flow 
through the wall of a pipe which carries a thick layer of insulation. Equa¬ 
tion 20:1 is applied to these situations below. 

Conduction of Heat through a Homogeneous Plane Wall. This case is 
illustrated in Fig. 20:1. Because of the homogeneity of the wall, the 
temperature gradient will be constant throughout its thickness if any 
change of k with temperature is neglected. Then dt/dx = ( ti — t 2 )/x, 
where t\ and t 2 are, respectively, the skin (surface) temperatures of the 
wall on its warmer and cooler sides and x is the total thickness of the wall. 
When k is in the units of Table 20:1, x is in inches and the temperature in 
degrees Fahrenheit. Equation 20:1 applies to a slice of the wall of 
infinitesimal thickness in the direction of heat flow. For the entire wall, 
the substitution suggested above gives 

Qcond — A — (t\ — t 2 ) — AC{t\ — t 2 ) — A (t\ — t 2 ) (20:4) 

in which the meaning of all terms except C and R has already been 
explained. C = (k/x) is the conductance of the wall; it represents the 
ability of the wall of thickness x to conduct heat and, using the system of 
units we have adopted, is expressed in Btu/(ft 2 )(°F)(hr). The reciprocal 
of the conductance is the resistance which the wall offers to the flow of 
heat, represented here by the symbol R.* 

A word of warning must be inserted here. The usual problem with 
which the engineer deals is the calculation of the heat flow between two 
fluids on opposite sides of a solid wall. He usually bases his solution of 
the problem on known or assumed values for the temperatures of these 
fluids at points removed from the wall surfaces, and the wall skin tem¬ 
peratures ti and t 2 cannot properly be assumed equal to the assigned fluid 

* Not, of course, to be confused with the gas constant, designated by the same 
symbol. 


THE TRANSMISSION OF HEAT 


497 


temperatures for reasons which have already been explained in our pre¬ 
liminary discussion of the effect of convection. We shall not be equipped 
for the solution of the engineering problem in its usual form until we have 
explored the effects of convection in greater detail; this will be done later 
in the present chapter. 

Example 20:44.. A brick building wall 12 in. thick has an inside surface tempera¬ 
ture of 56.3°F, an outside surface temperature of 3.7°F. It is 9 ft high and 20 ft long, 
(a) What is the heat transfer by conduction through this wall per hour? ( b ) What is 
its conductance? (c) Its resistance? 

Solution: 


k (Table 20:1) = 5; A = ( 9 ) (20) = 180; x = 12 

(a) Q = A ^ (ti - U) = 180 (56.3 - 3.7) = 3940 Btu/hr 

(b) C = - = 4 = 0-417 Btu/(ft 2 )(°F)(hr) 

x 1Z 

( c ) R = L = 2.4 (ft 2 )(°F)(hr)/Btu 

Conduction of Heat through a Laminated Plane Wall. This problem is 
illustrated in Fig. 20:2. The three laminations are numbered from the 
left in the figure, and each, separately, con¬ 
sists of a homogeneous plane wall. Thus the 
resistance of the first lamination is Xi/ki, of 
the second x 2 /k 2 , and of the third x 3 //c 3 . 

These resistances may be added to obtain the 
resistance to the flow of heat by conduction 
through the complete wall, or 


R = R\ -f- R 2 -T Rz — y- T 


X2 , 
k 2 


Xz 

kz 


or 


C = 4 = 


1_ 

R 


Xi 

/Cl ^ 


X2 

k 2 


Xz 
^ kz 


(20:5) 


( 20 : 6 ) 


The value of C obtained in this manner may 
be substituted in Eq. (20:4), with t\ and t 2 of 
that equation substituted as the extreme sur¬ 
face temperatures of the composite wall, to 
obtain the heat flow by conduction. Any 
number of laminations may be handled in 



Fig. 20:2. Heat transfer by 
conduction through a lami¬ 
nated plane wall. 

this manner. 


The rate of heat flow through each lamination is the same when steady- 
state conditions have been attained, and we may write 


Q = AC(h -U) = A - (h - h) = 

Xi 


A ^ {U 
x 2 


-h) = A - (h - U) 

Xz 







498 


BASIC ENGINEERING THERMODYNAMICS 


or 


ki/xi C 


If ti and U are known or assumed, U may be calculated from this propor¬ 
tion. By setting up similar proportions, all interior skin temperatures 
may be calculated and a line drawn to show the temperature gradient 
through each lamination, as in the figure. 

Sometimes a type of composite wall is employed in engineering con¬ 
struction in which the surfaces of one or more of the laminations is not 
planar, the lamination thus being of variable thickness; one of the most 
commonly met examples is the lath-and-plaster wall. For standard 
types of construction such as this, tests are made of the wall as a whole, 
and its conductance is reported in tables. Thus for the standard wall of 
gypsum plaster on wood lath, having a thickness of approximately f in., 
these tables give a conductance of 2.5. This is equivalent to the ratio 
k/x for this wall and may be substituted for it in Eqs. (20:4) to (20:6). 

Occasionally one of the inner laminations ma}^ consist of a fluid; this is 
especially true of building walls, often constructed with internal air spaces. 
Unless the air in this space is confined in small cells to prevent convection, 
as, for example, when the space is filled with mineral wool, convective 
effects will enter the problem and cause the temperature gradient to be 
greater in the fluid laminations near the boundary surfaces of the air 
space. The methods discussed above should properly be changed to take 
this factor into consideration. The accuracy required in the engineering 
calculation is often not extremely close, and sometimes the conductance 
of such air spaces is expressed in tables and used thereafter as k/x for the 
layer of air; these conductances will depend upon the thickness of the air 
space. This method is, of course, practical only for standardized types 
of construction but considerably simplifies the engineering calculation. 


Example 20:4 B. A building wall consists of fir siding and sheathing with a combined 
thickness of 1.62 in., 2- by 4-in. studs, and wood lath and plaster. The 3|-m. space 
between the lath and plaster and the sheathing is filled with mineral wool. Calculate 

(a) the conductance of this wall, (6) the heat flow per square foot by conduction when 
the inside surface of the plaster is at 67.3°F and the outside surface of the siding at 
0.7°F, and (c) the temperature in the center of the space that is filled with mineral 
wool. 

Solution: 


(a) C = 


1 

Xl , X2 , J_ 
ki + k 2 + C 3 


1.62 3.5 1 0-065 

0.8 + 0.27 + 2.5 


(b) Q = (1) (0.065) (67.3 - 0.7) = 4.34 Btu/(ft 2 )(hr) 

(c) t 2 = temp of inner surface of mineral wool; t 3 = temp of its outer surface 









THE TRANSMISSION OF HEAT 


499 


67.3 - 0.7 _ 67.3 - t 2 
2.5 0.065 

67.3 - 0.7 U ~ 0.7 


or t 2 = 65.7°F 


0.8/1.62 


0.065 


or t z = 9.5°F 


Averaging t 2 and tz to obtain the temperature at the center of the space, this tempera¬ 
ture is calculated as 37.6°F. 

Conduction of Heat through a Thick Homogeneous Cylindrical Wall. 
This case is illustrated in Fig. 20:3. The inner and outer surfaces have 
radii, respectively, of u and r 0 , the inner skin temperature being ti and the 
outer U. Assuming that t\ > U } 
the flow of heat is radially outward, 
and the area A for heat flow that 
enters into Eq. (20:4) changes as 
the flow proceeds from inner to 
outer wall; it is in handling this 
change in area that our problem 
differs from the two previously 
discussed. 

The entire wall may be divided 
into annular sections having an 
infinitesimal thickness dr such as 
that shown at a distance r from the 
center of the cylinder. The areas 
of the inside and outside surfaces of 
one of these annular segments, per 

unit length of cylinder axis, differ only infinitesimally, and the heat flow 
through each of them is the same under steady-state conditions; for 
this homogeneous wall k is also the same for each. Applying Eq. (20:1) 
to one of these annular segments over a unit length in the direction of 
the cylinder axis, 

kA dt k(2Tvr) dt 



Fig. 20:3. Heat transfer by conduction 
through a thick cylindrical wall. 


dr = 


Q 


Q 


or 


dr _ 2 irk 

T = Q 


dt 


in which dr has replaced dx of Eq. (20:1) and Q is the heat transfei pci 
unit length of cylinder per hour. Note that, for consistency, k is expressed 
in (Btu) (ft)/(ft 2 ) (°F) (hr). Integrating between limits suggested by the 
boundary conditions at the inner and outer suifaces, 



2irk 

~Q 


dt 











500 


BASIC ENGINEERING THERMODYNAMICS 


or 


and 


, r 0 2irk , , 

logc ll) 


Q = 3 [k in (Btu) (ft)/ (ft 2 ) (°F) (hr)] 


log, (r 0 /r % ) 


(20:7) 


It will be observed that, under the assumption made in attacking this 
problem (that ti > t 2 ), the heat flow from Eq. (20:7) is negative and the 
fluid within the tube constitutes the “thermodynamic system” to accord 
with conventions as to the sign of heat flow previously adopted. If the 

opposite is the case, it is necessary 
only to reverse the sign of Q. 

Conduction of Heat through a 
Laminated Cylindrical Wall. One of 
the principal engineering examples 
of radial heat flow through a cylin¬ 
drical wall is found when a hot or a 
cold fluid flows within a metal pipe 
which is surrounded by thick layers 
of insulation. This type of problem 
is illustrated in Fig. 20:4. The 
inner shell of material having a con¬ 
ductivity k i extends between radii r i 
and r 2 , and its surface temperatures 
are ti and t 2 , as shown. The ma¬ 
terial of the outer shell has a conductivity of k 2 and extends from r = r 2 
to r = r 3 , with surface temperatures similarly notated. Under steady- 
state conditions, the heat flow rate through both shells is the same, or 



Fig. 20:4. Heat transfer by conduction 
through a laminated cylindrical wall. 


q 2x k i (^2 t\) 2irk 2 (tz — t 2 ) 

log, (r 2 /Vi) log, (r 3 /r 2 ) 


or 


Then 


t 2 — t\ — 


Q log, {r 2 /r i) 

2irki 


and 


and tz — t 2 = 


tz — ti — (t 2 ~ ti) -f* (tz — t 2 ) = Q 

Ztt 


Q = 


Q l oge (r 3 /r 2 ) 
2irk 2 


loge (r 2 /ri) log, (r 3 /r 2 ) 
k i k 2 


2ir{tz — ti) 


1 ! r 2 ^ 1 , r 3 


( 20 : 8 ) 

















THE TRANSMISSION OF HEAT 


501 


This equation may be extended to apply to any number of cylindrical 
laminations in an obvious manner. Values of the conductivity are 
expressed in (Btu)(ft)/(ft 2 )(°F)(hr) in substituting in Eq. (20:8). 

Example 20:4 C. A steel pipe with an outside diameter of 4.5 in. and a wall thick¬ 
ness of 0.25 in. is covered with a 2-in. thickness of 85 per cent magnesia. The surface 
temperature on the inside of the pipe is 300°F, on the outside surface of the insulation 
is 90°F. (a) What is the heat flow per hour per foot of pipe length? (6) What is the 

temperature at the outer surface of the steel pipe? (c) Based on its inside surface 
area, what is the conductance of the pipe and insulation? 

Solution: 

(a) Let ki denote the conductivity of steel and k 2 that of the insulation. Also, n = 
inner radius of steel pipe, r 2 = its outer radius, and r 3 = outer radius of insulation. 
Then, from Eq. (20:8), 

n _ _ 6.28(90 - 300) -1321 

^ 12 2.25 12 4.25 ~ 0.0047 + 16.27 “ -81.2 Btu/(ft)(hr) 

300 ge 2.00 + 0.47 l0ge 2.25 


(6) Substituting this value of Q in Eq. 20:7 and applying it to the steel pipe, 

-81.2 = (6 -28 )(3Q 0)d2 - 300) Qr = 299 94 o F 
12 log e (2.25/2.00) 

(c) The inside diameter is 4 in. The inside area per foot of pipe length is 47r/12 
1.05 ft 2 . From Eq. (20:4), 


C = 


Q 


81.2 


A(h - U) 1.05(300 - 90) 


= 0.369 Btu/(ft 2 ) (°F) (hr) 


20:5. Heat Transfer by Radiation. Reviewing the earlier brief dis¬ 
cussion of the nature of radiated heat, it is noted that the following points 
were made: 

1. All bodies radiate heat in the form of an electromagnetic wave 
motion similar to that which carries light. 

2. The intensity of these waves (the amount of energy radiated from 
the body) is proportional to the fourth power of the absolute temperature 
of the surface from which they issue. 

3. A second body, intercepting some of these rays, may absorb, reflect, 
or transmit the energy carried along them. 

4. The perfect transmitter of radiant-heat energy is a void; the ability 
of matter, whether solid or fluid, to transmit the wave motion tends to 
increase with the transparency of the material body. 

Some elaboration is desirable at this point. The distinction between 
the wave that carries heat and that along which light is transmitted is 
found in its wave length. The heat waves occupy a band of wave lengths 
that lies between the wave lengths of light and the wave lengths of radio, 
being longer than the former, shorter than the radio waves; there is some 










502 


BASIC ENGINEERING THERMODYNAMICS 


overlapping of these bands, especially between the heat and the light 
waves. The heat wave obeys the same laws as the light wave, including 
the laws of reflection and refraction. 

The total amount of heat radiated from a body is a function not only of 
the fourth power of the absolute temperature of its surface but also of the 
area of its surface and of a property of the body called its emissivity, 
denoted by the symbol e. This emissivity may be changed by changing 
the nature of the surface of the body; in general, it may be increased by 
covering the surface with a dull black coating, as by the use of lampblack. 
The highest value that e may have is 1; this value is assigned to a hypo¬ 
thetical block body having maximum emissive properties. The emissivi- 
ties of nonblack bodies express their relative emission of radiant heat in 
comparison with that of a black body. The wave length corresponding 
to the heat-wave emanations from a black body is that corresponding to 
maximum intensity of heat transfer; it will vary inversely in length with 
the absolute temperature of the radiating surface. Emanations from 
bodies at higher temperatures are therefore characterized by shorter wave 
lengths, closer to those of the light band or even encroaching upon them. 

Of all the radiant-heat rays intercepted by a body, a part will be 
absorbed and will become, at least temporarily, stored thermal energy in 
that body; a part will be reflected into the space surrounding the body, 
and, based on the First Law, account must be made for the balance as 
having been transmitted through the body into the space beyond it. 
Most solids (an important engineering exception is glass in thin panes) 
permit the transmission of a negligible portion of the heat rays that fall on 
them. This is also true of liquids, but, for most gases, the portion trans¬ 
mitted is very large; thus the air of our atmosphere allows the major 
portion of the radiant heat from the sun to reach the solid surface 
of the earth. Opaque bodies having smooth, polished surfaces, light in 
color, are the best reflectors of the radiant-heat wave. The temperature 
of the surface of a mirror, held normal to the sun’s rays, is far below the 
surface temperature of a brick wall similarly placed with respect to those 
rays; since neither transmits an appreciable portion of the rays, much of 
the difference must be ascribable to the relative amounts of energy 
reflected and absorbed. 

Usually of greatest importance to the engineer, in making calculations 
that have to do with the transfer of heat by radiation, is the absorptivity 
a of the surface that intercepts the radiant-heat rays. The best absorber 
is the hypothetical black body, which has the highest emissivity, and the 
value of 1 is assigned to a for a black body, the absorptivity of nonblack 
bodies being compared with the standard so set. Since the black body 
absorbs all of the radiant-heat energy falling upon its surface, reflecting 
or transmitting none of the rays, a also represents the portion of the total 


THE TRANSMISSION OF HEAT 


503 


radiant-heat energy received which is absorbed. Exactly the same 
characteristics which make a surface a good emanator of radiant heat 
cause it to be a good absorber, and the value of a for the nonblack body is 
therefore equal to e for that body. Accordingly, a known value of either 
for a specific surface may be substituted for the other. This equality of a 
and e is based on Kirchhoff’s law. 

20:6. The Stefan-Boltzmann equation evaluates the heat radiated 
from a black body in terms of its radiating area (in the case of a smooth 
spherical body, this radiating area is the surface area) and its absolute 
temperature, 

Qrad = oAT* (20:9) 

where o' is a constant, known as Stefan’s constant, having a value of 
17.3 X 10 -10 when A is given in square feet, T in degrees Rankine, and 
Q rad in Btu per hour. 

The engineer is often less concerned with the total radiant-heat ema¬ 
nating from a body than he is with the portion of that heat which is 
absorbed by a second body, which happens to constitute the thermody¬ 
namic system on which his attention is centered. If that second body is 
totally enclosed within a radiating surface, it must absorb all of the energy 
which emanates from that surface. Therefore, no matter what the 
character of its surface, it becomes a black body in respect to its absorp¬ 
tivity. A totally enclosed body is a black body. 

In the meantime, of course, the enclosed body is the source of radiant- 
heat waves of its own, and, until a steady state is reached at which its 
temperature is equal to that of the surface which surrounds it, there will 
be a net heat exchange between the two. Remembering that the enclosed 
body is a black body, the Stefan-Boltzmann equation may be applied to 
measure this net heat exchange as 

= vAi(TA — TV) (20:10) 

in which the subscript 1 refers to the enclosed body, 2 to the surrounding 
surface. With respect to the sign of Q , the equation has here been written 
on the assumption that the enclosed body is the thermodynamic system. 

When the enclosure is not complete, all of the radiant heat from a radi¬ 
ating source may not be intercepted by the thermodynamic system for 
which the engineer desires to determine the net radiant-heat transfer. 
Nor are the emissivities and absorptivities of either body necessarily 
those of the black body, as in the case represented by Eq. (20:10). That 
equation must then be modified to take these factors into consideration; 
the new form is 

Q rad = uAiFeF c {TA ~ TV) (20:11) 

in which F e is an emissivity factor, depending on the emissivities of the two 


504 


BASIC ENGINEERING THERMODYNAMICS 


Table 20:2. Configuration and Emissivity Factors for Various Relations 

between Solid Surfaces* 

(Use in conjunction with Eq. 20:11. Subscript 1 refers to enclosed body) 


Item 

Surface relations 

Area 

used, 

A 

Configuration factor, 
F c 

Emissivity factor 
F t 

1 

Infinite parallel planes 

Either 

1 

1 

- + - - 1 
€i 62 

2 

Completely enclosed 

body without depres¬ 
sions and small com¬ 
pared with enclosure 

Ai 

1 

6] 

3 

Completely enclosed 

body, large compared 
with enclosure 

A\ 

1 

1 

- + ~ - 1 

6l 62 

4 

Concentric spheres or 
infinite cylinders 

A i 

1 

1 

- + yQ- - 0 

6i A 2 \C2 ' 

5 

Direct radiation between 
parallel and equal 
squares or disks of 
width D and distance 
apart L 

Either 

Fig. 20:5, curves 1 
and 2 

6\6l 

6 

Same as 5, except planes 
connected by noncon¬ 
ducting and reradiating 
walls (openings through 
furnace walls) 

Either 

Fig. 20:5, curves 3 
and 4 

6162 

7 

Same as 6 for slots, D = 
width 

Either 

Fig. 20:5, curve 4 

6l6 2 

8 

Two equal rectangles in 
parallel planes, directly 
opposite each other 

Either 

(Fc X F c ")* 

Fc = F c for squares 
(Fig. 20:5) equiva¬ 
lent to smaller side of 
rectangle. F " = F c 
for squares (Fig. 
20:5) equivalent to 
larger side of rectan¬ 
gle. Use curve 2 

If A is small com¬ 
pared with L, use 
eie 2 . If A is large 
compared with L, 
use F e , item 3 

9 

Two rectangles with a 
common side in per¬ 
pendicular planes 

Either 

Fig. 20:6 

6162 


* Adapted from Croft, as based on Hottel. 






























































THE TRANSMISSION OF HEAT 


505 


Table 20:3. Total Normal Emissivities and Absorptivities* 


Material 

Temperature of radiating 
or absorbing body, °F 

5 = solar radiation 

Emissivity e 
or 

absorptivity a 

Aluminum, oxidized. 

100-1000 

0.11-0.18 

Polished. 

100-1000-3000 

0.04-0.08-0.17 

Brick, building, cream. 

s 

0.38 

Red. 

s 

0.70 

Fire clay. 

2500 

0.75 

Refractory, black or chrome. 

200 -1000-2000 

0.92-0.97-0.98 

White or light buff. 

200 -1000-2000 

0.91-0.65-0.33 

Concrete. 

2500-5 

0.63-0.65 

Copper, polished. 

100 -1000-2000 

0.04-0.18-0.17 

Glass. 

100 

0.90 


100 

0.63 

Iron, galvanized, dirty. 

s 

0.89 

Limestone. 

100 

0.95 

Marble. 

100 

0.56 

Oak, planed. 

100 

0.91 

Paint, aluminum. 

100 

0.65 

Bronze. 

100 

0.51 

Smooth white. 

s 

0.20-0.40 

Black. 

200-600-1000 

0.92-0.95-0.97 

Red. 

200-600-1000 

0.95-0.95-0.85 

Green. 

200-600-1000 

0.93-0.90-0.80 

Paper, asbestos. 

100 

0.93 

Roofing. 

70 

0.91 

White. 

100-5 

0.80-0.28 

Pigments, lampblack. 

125-5 

0.96-0.97 

Black lacquer. 

125 

0.80 

White enamel. 

125 

0.92 

Dark varnish. 

125 

0.89 

Plaster, rough lime. 

100 

0.91 

Steel, polished. 

100-1000-5 

0.07-0.14-0.45 

Oxidized. 

100-1000 

0.79-0.79 

Water. 

100 

0.95 


* From Huber O. Croft, “Thermodynamics, Fluid Flow, and Heat Transmission, McGraw Hill 
Book Company, Inc., New York, 1938. 


bodies and, in some cases, their relative surface areas for radiation and 
absorption; F c is a configuration factor which expresses the effect of the 
relative position of the two surfaces on the average angle of incidence at 
which the thermodynamic system receives the radiant-heat waves of the 
other body. For an enclosed body, or for two parallel plane surfaces of 
infinite extent, F c is 1 since the rays are, on the average, normal to both 
surfaces. The maximum value which either of these two factors, F e or 
F c , may have is 1. Table 20:2, in combination with Figs. 20:5 and 20:6, 












































50G 


BASIC ENGINEERING THERMODYNAMICS 


gives the basis of calculation of F e and F c for radiating and absorbing solid 
surfaces disposed in various manners. Table 20:3 presents the emissivi- 



Fig. 20:5. Configuration factors F c for parallel surfaces (see Table 20:2). 



Fig. 20:6. Configuration factors F c for radiation between adjacent rectangles in per¬ 
pendicular planes. Y = ratio of length of unique side of area presented by the 
“thermodynamic system” divided by length of the side common to the two rectangles 
(=w/ 1). Z = ratio of length of unique side of other rectangle divided by length of 
the common side (=h/1 ). 

ties and absorptivities of various surfaces, showing, when appropriate, 
the variation of emissivity with surface temperature. 1 

In the engineering problem, the thermodynamic system is almost 
invariably the body at lower temperature. A case in point is the calcula- 

1 These tables and charts are adapted from Huber 0. Croft, “ Thermodynamics, 
Fluid Flow, and Heat Transmission,” McGraw-Hill Book Company, Inc., New York, 
1938. 







































































































THE TRANSMISSION OF HEAT 


507 


tion of the radiant heat received by a surface, such as a building wall, as 
the result of exposure to the rays of the sun. A surface held normal to 
the direct rays of the sun at the surface of the earth will, on a clear day, 
receive about 300 Btu/(ft 2 )(hr) in heat radiation from the sun. When 
the surface is inclined to the sun’s ravs, the amount of heat received is less 
and may be estimated by multiplying 300 by the sine of the average angle 
which the sun’s rays make with the surface. As has been explained 
above, a part of this heat will be reflected, a part transmitted, and the 
rest absorbed. For surfaces used in building construction, with the 
exception of clear panes of glass, the proportion transmitted is negligible; 
for clear glass panes, nearly all is transmitted. The proportion absorbed 
is, of course, the absorptivity (or emissivity, since the two are equal) of 
the surface; Table 20:3 shows, for appropriate surfaces, the emissivities 
under solar radiation. 


Example 20:6A. A rectangular 12- by 16-ft furnace has side walls 10 ft high. 
One of the 16-ft vertical walls is completely obscured by water tubes having a surface 
temperature of 500°F. The other side walls are of light buff refractory brick and 
have a surface temperature of 2200°F. Calculate the radiant heat received by the 
tube bank (a) from each of the adjacent vertical walls and ( b ) from the opposite verti¬ 
cal wall. 

Solution: 

(а) This corresponds to item 9 of Table 20:2. A\ = 16 X 10 = 160 ft 2 . From 

Table 20:3, ei (oxidized steel) = 0.79; e 2 = 0.33. Entering Fig. 20:6 with Y = y$ 
= 1.6 and Z = = 1.2, read F c = 0.155. F e = e x e 2 = (0.79)(0.33) = 0.261. 

Substituting in Eq. (20:11), 

Qrad = (17.3 X 10“ 10 ) (160) (0.261) (0.155) (2660 4 - 960 4 ) = 550,000 Btu/hr 
Total for the two adjacent side walls, 1,100,000 Btu/hr 

(б) Item 8, Table 20:2 applies. Ai = 160 ft 2 . Entering Fig. 20:5 with D/L = 

= 0.833, F c ' = 0.18 (curve 2). Entering this figure with D/L = -y| = 1.33, F c " = 
0.30 (curve 2). 

F c = (0.18 X 0.30)* = 0.232 


Area A is large compared with L. Therefore, from item 3, Table 20:2, 


F e 


1 

1/0.79 + 1/0.33 - 1 


0.303 


Substituting in Eq. (20:11), 

Q rad = (17.3 X 10- 10 ) (160) (0.303) (0.232) (2660 4 - 960 4 ) = 950,000 Btu/hr 

Example 20:6 B. Calculate the total radiant heat absorbed per hour by a building 
wall 100 ft 2 in area on which the sun’s rays fall at an angle of 70°. The exposed surface 
of the wall is of red brick. 

Solution: 


Qrad = (100)(300)(0.70) sin 70° = 20,400 Btu/hr 



508 


BASIC ENGINEERING THERMODYNAMICS 


20:7. Radiant-heat Transmission of Gases. The best medium for the 
transmission of radiant heat is a complete void. In discussing the trans¬ 
fer of radiant heat between solid surfaces, no attention has been given in 
most cases to the intervening gaseous material that always occupies at 
least a part of the space between those surfaces in practice. The data on 
solar radiation are an exception; the stated value of 300 Btu/(ft 2 )(hr) at 
the earth’s surface makes an allowance of 100 Btu/(ft 2 ) (hr) for radiant 
heat absorbed by the earth’s atmosphere as the sun’s rays pass through it, 
the total received at the outer surface of the atmosphere being some 400 
Btu/(ft 2 )(hr). Almost all of this difference is absorbed by the carbon 
dioxide and the water vapor in the atmosphere, the latter being the more 
important of the two in contributing to this effect. 

Heat transfer by radiation is also of maximum importance in the design 
of the furnaces used in industry to serve as the source of heat supply to the 
thermodynamic cycle. At least a part of the interior surface of these fur¬ 
naces is often covered with some refractory material which has the ability 
to resist destruction at high temperature. Other surfaces, such as banks 
of tubes, may be designed to transmit heat by conduction directly to the 
working fluid. These surfaces receive heat from the hot gases by conduc¬ 
tion, aided by convective effects, and by radiation; when a surface is “in 
sight” of the flames in the furnace, the transfer by radiation is usually 
much greater than by conduction. The refractory-covered surfaces are 
backed by thermal insulation so that almost none of the heat that reaches 
them escapes to the surrounding atmosphere; nearly all is either immedi¬ 
ately reflected back into the furnace spaces or is temporarily absorbed, 
creating a high surface temperature so that the refractory becomes a 
source of high-intensity radiant-heat waves of its own. This reradiation 
of heat back into the furnace spaces leads to higher temperatures in these 
spaces and thus to more complete combustion of the fuel. Above a cer¬ 
tain limit which is controlled by the characteristics of the fuel, the method 
of mixing fuel and air, and the method of ash disposal, a further increase 
of furnace temperature is not desirable. Furnace temperatures may be 
controlled by using surfaces around a part of the furnace periphery which 
are not refractory but pass the heat which they receive directly into the 
working fluid; an example is the tube bank to which reference has been 
made above. The heat which these metal tubes absorb is almost all 
passed directly on into the fluid that passes through them; they are thus 
kept at a level of temperature far below the firebox temperature, and 
their destruction is avoided. Because of their relatively low tempera¬ 
ture, they reradiate a negligible proportion of the heat which they receive 
either directly from the furnace gases or by reradiation from refractory 
furnace walls through those gases. Of the heat reradiated toward these 
cooler surfaces by refractory-covered furnace walls, a part will be absorbed 


THE TRANSMISSION OF HEAT 


509 


by the furnace gases; but this absorption raises the temperature of the 
gases and increases the intensity of the radiant waves they send out. 
The effect of absorption of this reradiated heat by these gases is thus 
largely discounted. 

The amount of heat received by furnace walls through direct radiation 
from the hot gases depends on the emissivities of those gases. For the 
purpose of studying these emissivities, these hot gases are classified under 
the following categories: 

1. Inactive gases, including nitrogen, excess oxygen, carbon dioxide, 
and water vapor, which are not undergoing a chemical change and which 
contain no suspended solids, such as particles of carbon or of fly ash. Of 
these, the only important radiators are the carbon dioxide and the water 
vapor. Significant emissivities are also associated with unburned hydro¬ 
carbons, carbon monoxide, and sulfur dioxide, but the percentages of 
these gases are low. 

2. Nonluminous flames, practically invisible in the furnace, which 
result from rapid combustion unaccompanied by cracking, or decomposi¬ 
tion, of the fuel. The emissivity of nonluminous flames varies from about 
0.1 to about 0.2. 

3. Luminous flames, caused by the burning of small particles of solid 
carbon as they are carried in suspension. These carbon particles often 
result from the cracking of hydrocarbons in the course of their combus¬ 
tion. The emissivity is much higher than for the nonluminous flame, 
averaging around 0.7. 

4. The gas mixture in the pulverized-coal furnace. This mixture 
includes not only particles of burning carbon but also unburned particles 
of solid fuel and particles of ash that have completed their combustion but 
are still held in suspension. This category has been given separate study 
because of its practical importance in the present-day power plant. 

The subject of heat transmission by radiation from gases and flames is 
far too extended to be discussed in detail in these pages, and the reader is 
referred for more comprehensive treatment to other sources. 1 

20:8. Effect of Convection on the Transmission of Heat by Conduction. 
Film Coefficients. In our initial discussion of the effects of convection on 
the amount of heat transfer by conduction (see Art. 20:1), it has developed 
that those effects result from an increase in the temperature gradient of 
the fluid in the layers close to the surface of a solid wall; this change from 
the uniform temperature gradient to be expected in the fluid if there had 
been no convection has been pictured in Fig. 20:1. In the usual case the 

1 See Huber O. Croft, “Thermodynamics, Fluid Flow, and Heat Transmission,” 
pp. 187-196, McGraw-Hill Book Company, Inc., New York, 1938; H. C. Hottel, 
Trans. ASME, 1935; Haslam and Hottel, Trans. ASME, 1928; Hottel and Mangels- 
dorf, Trans. Am. Inst. Chern. Engrs., 1936. 


510 


BASIC ENGINEERING THERMODYNAMICS 


temperatures of wall and fluid are low enough so that the heat flow by 
radiation between the two may be considered negligible; then it may be 
accepted that the heat transfer between wall surface and fluid is entirely 
by conduction. If the temperature gradient in the layers of fluid immedi¬ 
ately adjacent to the wall surface is known, either Eq. (20:1) or Eq. (20:4) 
may be applied to the calculation of the heat flow. When Eq. (20:4) is 
used, it is written in the form 

Qcond = A — (£2 — ts) 

x 

in which x is the thickness of a layer of the fluid having one of its surfaces 
in immediate contact with the wall surface and which is thin enough so 
that the temperature gradient is constant. The temperature t 2 is the 
common temperature of the wall surface and that surface of the fluid 
which is in contact with the wall; £3 is the temperature of the outer surface 
of the fluid layer. The constant temperature gradient through this thin 
film of fluid may then be substituted for (t 2 — tz)/x. 

This method of attack is of little use to the engineer, who usually begins 
his calculation of the amount of heat flow between two fluids with a 
knowledge of the fluid temperatures at points far enough removed from 
the wall separating the two so that the temperature gradient between 
these points and the wall surfaces with which the fluids are respectively in 
contact is not constant. The result of investigations into the effects of 
convection is therefore not reported in terms of the constant temperature 
gradient in the thin layer immediately adjacent to the wall but is expressed 
in terms of a film coefficient, h.* This film coefficient replaces k/x in Eq. 
(20:4) and the temperature of the fluid (U or t 0 ) replaces the intermediate 
(and unknown) temperature which appeared when the temperature 
gradient was substituted. It will be observed [see Eq. (20:4)] that this 
film coefficient is equivalent to the conductance C of a layer of fluid of 
just sufficient thickness to span the temperature interval between the 
skin temperature of the wall and the plateau level of fluid temperature t 0 
(or U ). Through a layer of this thickness the temperature gradient is not 
constant, and the value of h must take account not only of the layer thick¬ 
ness (depending largely on the viscosity of the fluid and the velocity of its 
flow) and its conductivity k but also the average temperature gradient. 

For some of the heat-transfer problems with which the engineer makes 
contact, the situation is more or less standardized, and it is feasible to 
determine the corresponding film coefficients by individual experiment 
and to publish them in tables for his use. For example, the film coeffi¬ 
cient often used for an inside building surface is 1.6, for outside surfaces 

* Not to be confused with specific enthalpy, though represented by the same symbol. 
The film coefficient is sometimes called a surface coefficient of heat transfer. 


THE TRANSMISSION OF HEAT 


511 


6.0. In both these cases the fluid is air, the larger value of the second 
coefficient being principally based on the expectation that wind on the 
outside of a building will give a forced-circulation effect; the specific wind 
velocity associated with the value of 6.0 which is cited above is 15 mph. 
Values of film coefficients vary greatly, ranging upward to over 1000 
Btu/(ft 2 )(°F)(hr) between wet steam and a cooler metal wall. 

The selection of a suitable film coefficient from a table is usually possible 
when the problem deals with ordinary forms of building construction. 
Here conditions are largely standardized since the fluid on both sides of 
the wall is air, the range of temperature is not great, and the kind, shape, 
and size of the solid materials which form the intervening wall are limited 
in the number of their variations. For the heat exchanger of the engi¬ 
neering power plant, on the other hand, the factors are more varied, and 
resort is usually made to a different method of approach, to be discussed 
later. When the film coefficient has been determined, the heat flow per 
hour may be based on the variation of Eq. (20:4) below, 


Q = hA(t f - t w ) (20:12) 

in which t f is the plateau level of fluid temperature at some distance from 
the wall and t w is the skin temperature of that surface of the wall with 
which the fluid is in contact. Of these two temperatures, tf is usually 
known to, or assumed by, the engineer as he enters upon the solution of 
the problem; t w is unknown, and Eq. (20:12) alone is not sufficient for a 
solution. However, a wall such as is pictured in Fig. 20:2 has, if the 
laminations are composed of solid material, two film coefficients. Let us 
designate that at the left-hand surface as E, that at the right-hand surface 
as h 0 . Since the heat flow through each lamination is the same and equal 
to the heat flow 7 into the left-hand, and away from the right-hand, surface, 
we may write 

hi(t, - tl) = - (<1 - k) = - (<2 - tz) = - (<3 - U) = hft 4 - to) 

X\ X 2 Xz 

Four independent equalities are expressed above, and since the only 
unknowns are the four skin temperatures t\, t 2 , tz, U, these temperatures 
may be established and the heat flow calculated. 

20:9. The Over-all Coefficient of Heat Transmission, U. It has been 
noted that the film coefficient h is equivalent to the conductance of the 
fluid layers adjacent to the surface of the solid wall. The corresponding 
resistance of these fluid layers is therefore l/h. In Fig. 20:2, the resist¬ 
ance of the fluid films may be added to the resistance of the wall to give 
the over-all resistance to heat flow between the two plateau levels of fluid 
temperature, U and t c . Thus 


512 


BASIC ENGINEERING THERMODYNAMICS 


and 


hi ki ' k 2 kz h 0 


R (20:13) 

hi ki'k 2 ' kz h 0 

Q = AU{U - to) (20:14) 


in which £/ is the over-all conductance of the entire wall, including the 
effect of the resistance offered by the fluid films, and is called the over-all 
heat-transfer coefficient. When the situation is sufficiently standardized, 
values of U may be tabulated for various walls; these simplify the 
engineering calculation of the heat flow since the value of all terms on the 
right side of Eq. (20:14) are then usually readily and directly available to 
the engineer. 

For reasons already explained, the practice of tabulating values of U 
for the engineer’s use is commonest in the case of a building wall or other 
building surface. When one of the surfaces is an outside surface, it is 
customary to assume a wind velocity of 15 mph. Also, when a wall con¬ 
tains an interior air space, two additional inside-surface coefficients are 
often used in computing U. Tabulations of the values of U are also used 
in a few other engineering situations where heat-exchanger design has 
assumed some semblance of standardization. For example, the tubes of 
surface types of steam condensers are largely standardized as to size and 
material, the fluids are always the same, and the temperature variation is 
small; this makes it feasible to supply information to the engineer in the 
form of charts showing the value of U for condenser tubes according to 
their size and the velocity of water flow through them. 

9 

Example 20:9 A. Assuming a wind velocity on the outside surface of 15 mph and 
using the film coefficients suggested in Art. 20:7, calculate the plateau temperatures 
ti and t 0 and the over-all heat-transmission coefficient U for (a) the wall of Example 
20:44. and ( b ) the wall of Example 20:45. (c) If the interior air space of the wall of 

Example 20:4 B had not been filled with mineral wool, compute U by using two addi¬ 
tional inside-surface coefficients, (d) What effect does the mineral wool have on the 
over-all conductivity of the wall? 

Solution: 


(a) From the solution of Example 20:44, Q = 3940 Btu/hr. Applying Eq. (20:12), 
3940 = (1.6) (180) (ti - 56.3) or U = 70°F 
and 


3940 = (6.0) (180) (3.7 - t 0 ) or t 0 = 0°F 


U = 


1 

1.6 ^ 5 ^ 6 


0.313 





THE TRANSMISSION OF HEAT 


513 


Checking, 


Q = AU(U - t 0 ) = (180)(0.313)(70 - 0) = 3940 Btu/hr 
(6) For the wall of Example 20:415, 


4.34 = (1.6)(l)(*i - 67.3) or U = 70°F 
4.34 = (6.0) (1) (0.7 - t 0 ) or t a = 0°F 

U = —-—-— = 0.062 

L6 + 0.065 + 6 

Checking, 


Q = (1) (0.062) (70 - 0) = 4.34 Btu/hr 

(C> U = J_+J_ + J_ + J_ + L6? + l 

1.6 ^ 2.5 ^ 1.6 ^ 1.6 ^ 0.80 ^ 6 


= 0.224 


(d) The use of mineral wool reduces the over-all conductivity of the wall by 


( 


0.224 - 0.062 
0.224 



72 per cent 


Example 20:9 B. In Example 20:4 C, assume the inside coefficient is 1000 and the 
outside coefficient is 2 Btu/(ft 2 )(°F)(hr). (a) Assume that the pipe contains wet 

steam, and calculate its pressure. ( b ) What is the ambient temperature of the air 
surrounding the pipe? (c) What is the over-all heat-transfer coefficient of the pipe 
and insulation, based on area of inside surface? 

Solution: 


(a) The inside surface area, per foot of pipe length, is 1.05 ft 2 . Based on the solution 
of Example 20:4 C, 

81.2 = (1.05) (1000) (k - 300) and U = 300.08°F 

For saturated steam, this temperature is equivalent to a pressure of 67.1 psia. 

(b) The ouside area of the insulation, per foot of pipe length, is 8.57r/12 = 2.23 ft 2 . 

81.2 = (2.23) (2) (90 - t 0 ) or t 0 = 71.8°F 

/ \ tj ___ Q_ _ _ _ 81.2 _ _ rv oqq 

{C) ~ A(U - to) ~ 1.05(300.08 - 71.8) 

20:10. The Application of Dimensional Analysis. Tables and charts 
giving the values of h or U may allow for only a limited number of variable 
factors. The total number of such factors that may play a part in heat 
transmission, when aided by convection, is large, as has been suggested in 
Art. 20:1. For the study of the less standard situation, dimensional 
analysis, to show the degree to which each of these factors enters, suggests 
itself. 

Following the procedure of dimensional analysis (see Art. 17:3), we 
first decide upon the factors which are important in their influence on the 














514 


BASIC ENGINEERING THERMODYNAMICS 


value of the film coefficient h. (Note that, dimensionally, [h] = [M/6 3 T].) 

Let us suppose that these factors include: 

D = dimension of cross section of conduit through which fluid moves, 

as, for example, diameter of circular tube; dimensional^, [D] 

= [L] 

r m 

p = mass density of fluid; dimensionally, [p] = ji 


k = conductivity of fluid; dimensionally, [k] 


ML 

d*T 


V = velocity of fluid; dimensionally, [V] 


L 

d 


p = absolute viscosity of fluid; dimensionally, [p] 


M 

Ld 


c p = specific heat at constant pressure of fluid, dimensionally, 

= \ L 2 

J2 T 

L = length of conduit; dimensionally, [L] = [L] 

The method of Art. 17:3 may now be followed to give 



or 



(20:15) 


in which C is a dimensionless constant and a, b , and c are exponents the 
values of which must be determined by experiment. The group hD/k is 
dimensionless and is known as the Nusselt number (Nu). The dimen¬ 
sionless group pVD/p we recognize as the Reynolds number (Re). The 
group c p p/k, also dimensionless, is called the Prandtl number (Pr).* The 
value of the exponent c has been shown to be close to zero by experiment, 
and Eq. (20:15) is usually written in the form 

Nu = C(Re) a (Pr) b (20:16) 

The values of h, of D, and of k must be in consistent units if the Nusselt 
number is to be dimensionless. Thus, for h in Btu/(ft 2 )(°F)(hr) and D 
in feet, the unit of k must be (Btu)(ft)/(ft 2 )(°F)(hr). The same con¬ 
sistency must be observed with regard to the units of the quantities of 
which the Reynolds and Prandtl numbers are composed. For the Prandtl 
number, the conventional unit of p is the lb-sec/ft 2 and of c p the Btu/ 


* Based on the varying selection of the important factors, other dimensionless 
numbers, e.g., the Stanton, Peclet, and Grashof, also are found in the literature. 


















THE TRANSMISSION OF HEAT 


515 


(lb mass) (°F). If, as in the Nusselt number, the unit of k is the (Btu) (ft)/ 
(ft 2 )(°F)(hr), it is necessary, in making the time unit consistent, to multi¬ 
ply the conventional unit of viscosity by 3600 and, in making the mass 
units agree, to multiply the conventional unit of specific heat by g, or 
32.2. Thus Pr = (3600)(32.2)(12 )c P n/k = l,390,000c p ^/fc with c p and /j, in 
the conventional units suggested above and k in (Btu)(in.)/(ft 2 )(°F)(hr), 
the terms in which its values are tabulated in Table 20:1. 

The Prandtl number is composed entirely of properties of the fluid and 
therefore is itself a property. For gases, and for vapors at pressures 
approximating atmospheric pressure, its value is close to 0.75. For 
vapors at relatively high pressure, its value is higher and tends to vary 
more widely with temperature. Liquids also show higher values of Pr 
and a larger variation with temperature; the variation with pressure is 
small. 

That the effects of convection should vary with the Reynolds number 
is logical. We have seen in Chap. 17 that the Reynolds number not only 
establishes the point of transition from laminar to turbulent flow but also, 
in the higher range of values, is a key to the degree of turbulence. As the 
flow becomes more turbulent, the boundary layer becomes thinner and 
the temperature gradient in this layer greater, resulting in a higher rate of 
heat transfer. 

The factors which have been selected earlier in this article and which 
have led, in dimensional analysis, to the development of Eq. (20:16) have 
been based on the assumption that the convection is forced, the velocity 
of fluid flow being primarily controlled by external agencies, such as 
pumps or fans, rather than being due solely to the differences in fluid 
density which are caused by temperature differences, as in natural con¬ 
vection. Under forced convection, the flow almost invariably lies in the 
turbulent range. The application of Eq. (20:16) is therefore primarily to 
forced-convection problems, and the values of Re which apply are usually 
relatively high. If natural convection is accompanied by a Reynolds 
number in the turbulent range, this relation may also be used. 

The Nusselt equation (20:16) supplies a common pattern which can be 
followed by investigators of heat transmission in turbulent fluid flow in 
expressing the results of their experiments. When the fluid flows within 
a pipe round in cross section, as is often the case in the engineering prob¬ 
lem, the problem is of a more standardized character and, as would be 
expected, closer agreement is found between results as reported by various 
experimenters. It is not within the scope of this text to list and compare 
the various equations that have been suggested. It will suffice for our 
purposes to say that, for turbulent flow inside a round pipe, the consensus 
seems to place the value of the exponent a in Eq. (20:16) at about 0.8, of 
b at 0.4, and of the exponent c in Eq. (20:15), when the D/L ratio is 


516 


BASIC ENGINEERING THERMODYNAMICS 


included, at 0.05. There is less consistency in the values suggested for 
the dimensionless constant C, the recommended values falling in the 
approximate range 0.023 (when the ratio D/L is omitted) to 0.032 (with 
c = 0.05). 

The matter of whether the fluid is being heated or cooled will affect 
the amount of heat transfer and thus the size of the film coefficient, espe¬ 
cially if the velocity of fluid flow is low. When the fluid receives heat, 
the layers of fluid adjacent to the solid wall from which the heat enters 
will be at higher temperature than the bulk of the fluid. If the fluid is a 
liquid, this higher temperature will mean that the fluid is less viscous near 
the wall and the parabolic velocity profile (see Chap. 17) is flattened, the 
ratio of average to maximum fluid velocity being increased. If a gas, 
increased temperature means increased viscosity and the layers of gas 
near the wall will be slowed, elongating the velocity profile and decreasing 
the ratio of average to maximum fluid velocity. Film-coefficient formu¬ 
las often make an allowance for this effect by using different coefficients 
C for heating and for cooling. The difference is more pronounced for 
liquids than for gases, since the viscosity of liquids is more sensitive to 
temperature change. At the higher Reynolds numbers associated with 
forced convection, the effect is ordinarily negligible. 

As the basis of the solution of exemplary problems, we shall select a 
comparatively simple relation to apply to the turbulent flow of fluids 
inside of round pipes, 

hD 

= 0.023 (Re) °- 8 (Pr) °- 4 * (20:17) 


in which the meaning of all terms has previously been discussed. Note 
that hD/k is the Nusselt number and the units of h, D, and k must be 
selected so that it will be dimensionless. The equation is valid for 
Reynolds numbers above 3000; it may be applied to noncircular ducts 
with D equal to four times the hydraulic radius. 


Example 20:10A. Air at atmospheric pressure and 120°F moves through a 10- by 
16-in. rectangular steel duct with an average velocity of 20 fps. Calculate the inside 
film coefficient. 

Solution: 


Hydraulic radius = 


cross-sectional area 
wetted perimeter 


D = 



1.02 ft; 


A 

12 



[1 


(10) (16) 
2(10 + 16) 


3.07 in. 


+ (0.00165) (120 - 32)] = 0.0155 


[Table 20:1] 

c p = 0.24 Btu/(lb mass)(°F) 

m = [0.0165 + 2.5 X 10 -5 (120)]/47,800 = 4.07 X 10~ 7 lb-sec/ft 2 [Table 17:1] 


* W. H. McAdams, “Heat Transmission,” 2d ed., McGraw-Hill Book Company, 
Inc., 1942. 






THE TRANSMISSION OF HEAT 


517 


p = 


P _ (14.7)(144) 

gRT (32.2) (53.3) (580) 


= 0.002125 slug/ft 2 


Pr = 1,390,000 -fp = 

K 


c P n _ (1,390,000) (0.24) (4.07) (10 -7 ) 


= 0.73; (Pr) 0 - 4 = 0.882 


(12) (0.0155) 

0 pVD (0.002125)(20)(1.02 X 10 7 ) ^ ^ 

Re = —— = ^^- L = 106,000; (Re) 0 - 8 = 10,400 

Substituting in Eq. (20:17), 

Nu = = (0.023) (0.882) (10,400) = 211 

h = (211) 1 ( o 2 155> = 3-2 Btu/(ft 2 )(°F) (hr) 


The use of the value of 0.75 suggested in the text for Pr of gases will give (Pr) 0 - 4 = 
0.891 and change h in this problem to 3.24, a negligible difference. 


When the fluid flows turbulently through an irregularly shaped con¬ 
duit, as over the outside surfaces of a bank of tubes, the problem assumes 
a much less standardized pattern; for example, the selection of a repre¬ 
sentative dimension D is less readily made. Experiments usually are 
restricted to a narrower field of investigation, and results are sometimes 
reported in a form which is hardly recognizable as conforming to the basic 
equation (20:16). A complete survey is inappropriate to our discussion, 
and only one simple example will be given here of such equations. This 
is suggested by Still 1 and applies to the flow of gases normal to banks of 
tubes. For cooling of the gas, he gives 

Nu = 0.279F(Re) 0 - 6 (20:18) 

and, for heating, 

Nu = 0.29F(Re) 0 - 6 (20:19) 

In both of these equations F is a constant which depends on the number 
of rows of tubes in the direction of gas flow and their arrangement. If 
the rows are staggered, F varies from 1 for two rows of tubes to 1.335 for 
six rows; if the rows are in line, from 0.835 for two to 0.905 for six rows of 
tubes. It will be observed that the Prandtl number does not appear in 
these equations. However, if the fluid is a gas as specified, the Prandtl 
number has a nearly constant value which, raised to a suitable constant 
power, may be included in the constant. The dimension D in the Nusselt 
number refers to the outside diameter of the tubes, and k is based on the 
average temperature of the surface in using these equations. In the 
Reynolds number, V is the gas velocity through the most constricted sec¬ 
tion, and both p and p are taken at the average temperature of the gas and 
the tube surface. 


1 Proc. Inst. Mech. Engrs., 134 (1936). 








518 


BASIC ENGINEERING THERMODYNAMICS 


Example 20:105. In the back pass of a boiler, furnace gases at 600°F flow across 
a bank of six staggered rows of 3-in.-outside-diameter tubes with a maximum velocity 
of 8 fps. The surface temperature of the tubes averages 400°F. Calculate the 
outside film coefficient for these tubes. 

Solution: 


D = A = 0.25 

The values for air will represent k, p, and p. for the hot gases, k is calculated at 400°F> 
M and p at (400 + 600)/2 = 500°F. 


A 

12 

M 

P 

Re 


0.163 

12 


[1 + 0.00165(400 - 32)] = 0.0218 


[0.0165 + (2.5 X 10- 5 ) (500)1/47,800 = 6.06 X 10' 
(14.7) (144) 


(32.2) (53.3) (960) 
(0.001285) (8) (0.25) (10 7 ) 
6.06 


= 0.001285 

= 4250; (Re) 0 - 6 = 150 


For cooling of the gas, use Eq. (20:18); for six staggered rows of tubes, F = 1.335. 


Nu = = (0.279) (1.335) (150) = 55.8 

k 

h = (55 ' 8 1 ( S?; Q218) = 4.86 Btu/(ft 2 )(°F)(hr) 
0.25 


20:11. Calculation of Film Coefficients—Laminar Flow. Under forced 

convection, laminar-flow conditions seldom exist. When they do, as 
sometimes in heat exchangers used in the refinement of viscous crude oils, 
Eq. (20:15) may be used as the basis for reporting the results of experi¬ 
ments; the value of the exponent a of the Reynolds number is usually 
reported as much smaller than the 0.8 of Eq. (20:17), which applies to 
turbulent flow. 

But laminar flow usually is the result of natural convection. To 
attack the problem dimensionally requires that we add to the list of fac¬ 
tors given in Art. 20:10, and which resulted in Eq. (20:16), additional 
factors which would assume importance in natural convection. These 
would include AT, the difference between the temperature of the main 
body of the fluid (its plateau level of temperature) and the temperature 
of the wall surface; (3, the coefficient of cubical expansion of the fluid, 
reflecting the effect of temperature changes on its density; and g, the 
acceleration of gravity. Other factors may, in special situations, be 
added to this list; for example, when the rate of change of viscosity with 
temperature is large, two viscosity factors may be introduced, the vis¬ 
cosity at the wall and the viscosity in the main body of the fluid. 

For each additional factor that enters into the dimensional analysis, 
the number of dimensionless quantities appearing in the final result will 
be increased by one. Analysis or experiment may show the effect of many 
of these numbers to be slight, as was the case with the D/L ratio of Eq. 







THE TRANSMISSION OF HEAT 


519 


(20:15) in turbulent flow; as a result they may be dropped from the final 
relation so that its application may be simplified. In natural convection, 
the important dimensionless numbers are, in the general situation, the 
Prandtl number, previously discussed, and the Grashof number (Gr), 
defined as 


Qi _ fig? AT D* 

M 2 


( 20 : 20 ) 


The meaning of all the terms has already been explained. Care must be 
observed to preserve the dimensionless character of the Grashof number 
in choosing the units of (3, g, p, AT, D, and p, but this is simpler than for 
the Prandtl number since the time period is the second in the conventional 
unit for all and the length unit is the foot. It will be observed that the 
Grashof number is not, like the Prandtl number, a property of the fluid 
but is rather to be compared with the Reynolds number. 

When natural convection takes place on a vertical surface, or one 
inclined to the horizontal, whose vertical dimension is large, the dimen¬ 
sionless ratio D/L assumes greater relative importance. Assembling the 
important dimensionless numbers into an equation of the form of Eq. 
(20:16), but which will apply to natural convection, we have 

Nu = C(Pr) b (~jr) (Gr)'* (20:21) 

in which C is again a dimensionless constant and b, c, and d are exponents. 
The values of all must be determined by experiment and may vary with 
the situation; for example, if the vertical dimension of the surface is small, 
the exponent c will approach zero in value, as it did for forced convection. 
As applied to horizontal cylinders (pipes), Nusselt determined the values 
of C = 0.99, b = d = i, and c = 0. Thus, for horizontal pipes, 

Nu = 0.99(Pr)*(Gr)* (20:22) 

For long vertical cylinders, Heilman 1 suggests that the coefficient C 
be changed to 1.2 to allow for the higher velocities created by the cumula¬ 
tive effect of convection on a vertical pipe. This increase in C takes the 
place of an inclusion of the ratio D/L in the equation. Thus, for long, 
vertical tubes, 

Nu = 1.2(Pr)*(Gr)* (20:23) 

Equations (20:22) and (20:23) apply to the calculation of the film 
coefficient on either the outside or the inside surface of a tube and to both 
liquids and gases, provided only that the thickness of the film depends on 

1 Trans. ASME, Fuels and Steam Power, 51(41), 297. 



520 


BASIC ENGINEERING THERMODYNAMICS 


natural-convection effects. The values of k, p, p, and c p used in com¬ 
puting the Prandtl and Grashof numbers are based on the mean tempera¬ 
ture of the film. 

Example 20:11 A. A 4.5-in.-outside-diameter horizontal pipe carries warm water 
through a room in which the ambient temperature is 70°F. The temperature of the 
outside surface of the pipe is 210°F. Calculate the outside film coefficient of the pipe. 

Solution: 


D = 4.5/12 = 0.375; AT = 210 - 70 = 140; g = 32.2 


/3, p, and p will be based on the average temperature of the film, or 140°F. k will be 
based on the surface temperature of the pipe, 210°F. is the rate of change of air vol¬ 
ume with temperature, which, at constant pressure, is 1/7 7 ; thus/3 = goo = 0.00167. 


p = 


(14.7) (144) 


= 0.00206 


(32.2) (53.3) (600) 
p = [0.0165 + (2.5 X 10- 6 )(140)]/47,800 = 4.18 X 10~ 7 
Gr = (0-00167)(32.2)(0.0(^06) 2 (140)(0.375) 3 (10) 14 _ 9>670 000; (Qr)i _ 55 7 

Assume Pr = 0.75. Then (Pr)* = 0.931. 
k 0.163 


12 


12 


[1 + 0.00165(210 - 32)] = 0.0176 


Nu = = (0.99) (0.931) (55.7) = 51.3 

h = (51,3 Q ) ff^ 7 — = 2.41 Btu/(ft 2 )(°F)(hr) 


A special situation of interest to the engineer is represented by the 
steam condenser, so frequently a part of the equipment of the steam power 
plant. In the surface type of steam condenser, an essentially quiet body 
of exhaust steam surrounds banks of tubes, usually horizontal, through 
which cooling water is forced. The film coefficient on the outside of these 
tubes is based on natural convection but is affected by the film of con¬ 
densate which forms on the tube surface and through which the heat must 
pass. Occasionally, when the tubes are new and the outside surface is 
polished, or when coated with a thin film of oil, the condensation may 
form as droplets. Drop condensation produces much larger film coeffi¬ 
cients, but since the surface of the tube soon becomes oxidized in use and 
this is favorable to the formation of a film, it is usually assumed for the 
purposes of design that film condensation applies. Nusselt 1 derived an 
equation to apply to film condensation, as follows: 


h = 5.62 



(20:24) 


1 Z. Ver. Deut. Ing. y 60, 541, 569 (1916). 









THE TRANSMISSION OF HEAT 


521 


in which h 

k 


P 


9 

hfo 

D 


P 

AT 


film coefficient, Btu/(ft 2 )(°F)(hr) 

conductivity of liquid condensate, (Btu)(ft)/(ft 2 )(°F)(hr) 

mass density of condensate, slugs/(ft 3 ) 

acceleration of gravity, ft/sec 2 

enthalpy of vaporization of vapor, Btu/lb 

diameter of tube, ft 

absolute viscosity of the condensate, lb-sec/ft 2 
temperature difference between vapor and the tube surface, 
°F 


It will be observed that the product pg is the weight density in pounds per 
cubic foot and that the square of the weight density of the liquid may 
therefore be substituted for p 2 g 2 in the equation. The temperature at 
which k, p, and p are calculated is the average of the temperatures of the 
wall and the steam; hf g at steam temperature is used. The value of h 
obtained is large, usually above 1000, and the equation is often by-passed 
in favor of the use of an assumed value of 1000; this practice is considered 
conservative in steam-condenser design. Even when h is as high as 2000, 
a relatively small effect will be produced on U, the over-all heat-trans¬ 
mission coefficient of the tube, by using the smaller value. 


Example 20-11 B. The water vapor in a condenser has a temperature of 100°F- 
The condenser tubes are 1 in. outside diameter and have an average outside-surface 
temperature of 80°F. Calculate the average outside film coefficient for these tubes. 
Solution: 

D = = 0.0833; hf g (at 100°F) = 1037.2. Jc, p, and p are calculated for water 

at the average between steam and surface temperature, or 90°F. 


12 = (it) [1 + °- 001 ( 90 ~ 68 d = 0-35 
pQ = 0.01610 = 62J lb/ft ' 


p = [2.317 
h = 5.62 [ 


- (0.0164)(90)]/47,800 = 1.75 X 10 
(0.35) 3 (62.1) 2 (1037.2)(10 5 ) 


(0.0833) (1.75) (20) 




5.62(59 X 10 8 ) 


l 

4 


= 1560 Btu/(ft 2 )(°F)(hr) 


20:12. Mean Temperature Difference. When the temperature of one, 
or both, of the fluids that exchange heat as they pass through the heat 
exchanger changes between entrance to and exit from that unit, a knowl¬ 
edge of the mean, or average, temperature difference between these fluids 
is necessary if equations such as (20:14) are to be applied to the calcula¬ 
tion of the heat given up by the one and received by the other in the course 
of their entire flows through the unit. This mean temperature difference 
can then be substituted for U — t 0 in Eq. (20:14). One of the simplest 
examples is supplied by the steam condenser, where one of the fluids 
remains at constant temperature while the other increases in temperature 





522 


BASIC ENGINEERING THERMODYNAMICS 



Fig. 20:7. Mean temperature differ¬ 
ence—condensation of a vapor. 


from inlet to outlet; it is illustrated in Fig. 20:7. The area of the solid 
wall which separates the fluids is represented along the abscissa of this 

figure, and it is assumed that each 
unit area of this wall has the same 
over-all heat-transmission coefficient 
U ; it is also assumed that heat leaving 
the hot fluid exactly equals that en¬ 
tering the cold. The constant tem¬ 
perature of the warmer fluid is desig¬ 
nated as l a , the entering temperature 
of the cooler as ti, and its exit tem¬ 
perature as £ 2 . The direction of flow 
of the cooler fluid is indicated by the 
arrow. 

The temperature difference t a — t\ 
is greater near the entrance to the 
heat exchanger than at the exit, where 
it is t a — U‘ The cold fluid therefore receives heat at a more rapid rate 
per unit of wall area near the entrance and will increase in temperature 
more rapidly than as the exit is approached; the solid curve 1-2 will 
approximate the temperature vari¬ 
ation with surface area traversed. 

There are two more general cases, 
for each of which the temperature 
of the hot fluid does not remain 
constant but decreases from t a to t b 
as it gives up heat. The first of 
these is illustrated in Fig. 20:8 and 
represents the temperature varia¬ 
tion of the fluids as they pass 
through a parallel-flow heat ex¬ 
changer. The two fluids enter at 
the same end of the heat exchanger 
and approach each other in tem¬ 
perature as they flow through it; 
the final temperature of the cold 
fluid is always less than the tem¬ 
perature at which the hot fluid leaves the exchanger. The difference 
between the temperatures of the two fluids at entrance is always greater 
than at exit, and the temperature variation is shown by the two curves, 
ah and 1-2, which bend toward each other. 

In approaching the problem of calculating the mean temperature differ¬ 
ence between the fluids for the entire heat exchange, let us consider a sec- 



Fig. 20:8. Variation of fluid temperature 
—parallel-flow heat exchanger. 















THE TRANSMISSION OF HEAT 


523 


; 


tion of the heater having a surface of the elementary area dA. The hot 
fluid enters this section of the heater at a rate which we shall designate as 
M h lb/hr and a temperature of fa; its temperature is fa + dfa (where dfa 
is negative) as it leaves the heater section. Designating the specific heat 
of the hot fluid as Ch, we may write 

dQh — MhCh dfa (1) 

in which dQh is the heat leaving the hot fluid as it passes over the elemen¬ 
tary surface area dA ; it is negative in sign according to the convention w r e 
have adopted. Similarly, for the cold fluid, 

dQ c = M c Cc dt c (2) 


in which the meaning of all terms is the same as in (1) but the subscript c 
refers to the cold fluid; since A t c is positive, dQ c must be positive. dQh 
and dQc are equal but opposite in sign, and therefore 

dQh = M h c h dth = —M c c c dt c (3) 

or 

dk = dQh (ij and - du = dQh (ib) (4) 

The difference, dfa — dt c , is the amount by which the difference between 
the temperatures of the two fluids changes, or 

dt h - dtc = d(tn - Q = dQh + jfAj ( 5 ) 

The difference in temperature fa — t c controls the rate of heat flow between 
the two fluids through the surface area dA. According to Eq. (20:14), 

dQh = —U(fa — t c ) dA ( 6 ) 


Substituting this value of dQ h in (5) and rearranging, 


d(fa fa ) 
fa t c 


= -U 




dA 



The left side of this equation will be integrated between the limits set by 
the temperature difference fa — fa at entrance, which will be designated as 
At', and that at exit, fa - fa, designated as At". The limits for the inte¬ 
gration of the right side are 0 and A. Then 



dQh tc ) 

fa tc 


= — U 


1 


+ 


1 


M h Ch M c c C/ 




or 


log, 


At 


n 


At' 


= — UA 




( 9 ) 














524 


BASIC ENGINEERING THERMODYNAMICS 


But 


or 


r m 


Q h = -UA (A0, 

Q h M h c h (t b — to) _ M c Cc(ti — < 2 ) 


-UA = 


(At) 


( 10 ) 


(A t) m (At) m 

in which (At) m is the mean temperature difference through the entire heat 
exchange. Substituting this value of — UA in (9), we obtain 

t b — t a ~\r t\ — _ At" — At 

(A t) m 


log. 


At 


At' 


or 


(A t) m = 


(At) 


At" - At' 


(ID 


( 12 ) 


loge (At"/At') 

It will be noted that the positions of At' and At" may be interchanged in 
this equation without affecting the value of (A t) m . (At) m is called the 
logarithmic mean temperature difference (LMTD) and therefore 

At" - At' 


LMTD = 


log e (A t"/At') 


(20:25) 


in which At" is the temperature difference between the fluids at either end 
of the heat exchanger, At' being the difference at the other end. 

Figure 20:9 illustrates the counter¬ 
flow principle in heat-exchanger de¬ 
sign in which the fluids enter at oppo¬ 
site ends of the heat exchanger and 
flow in opposite directions through 
the unit. Note that, in counterflow, 
the final temperature of the cold fluid 
may exceed the final temperature of 
the fluid which gives up heat and a 
lower average temperature difference 
is possible. It may be shown that 
Eq. (20:25) will apply as well to the 
calculation of the mean temperature 
difference for this situation (and to 
the conditions illustrated in Fig. 20:7) 
as for the parallel flow on which its 
derivation was based. If M h Ch = M c c c 
in counterflow, the temperature differ¬ 
ence between the fluids will remain constant. Equation (20:25) gives 
an indeterminate value for this special condition, but the arithmetical 
mean temperature difference (AMTD) may be used. This is a simple 



Fig. 20:9. Variation of fluid tempera¬ 
ture—counterflow heat exchanger. 
















THE TRANSMISSION OF HEAT 


525 


arithmetical average between At' and At". For the cases illustrated in 
Figs. 20:7 and 20:8, Eq. (20:25) should always be used, for the AMTD 
assumes that the temperature of the fluid follows a straight line between 
its initial and final values (see the dashed line 1-2 of Fig. 20:7); the calcu¬ 
lation of the AMTD will therefore show a value higher than the true mean 
temperature difference. In counterflow design, At' and At", even when 
not equal, may be so nearly the same that difficulty will be experienced in 
evaluating LMTD by Eq. (20:25). In that case, it may be better to use 
the AMTD. When At' and At" are equal, the LMTD and the AMTD are, 
of course, identical values; this may occur not only in counterflow but 
also when heat is exchanged between fluids one of which is condensing, 
the other evaporating, at respectively constant temperatures. The 
counterflow principle makes it possible, in theorjq to design a heat 
exchanger to operate on an infinitesimal temperature difference and thus 
avoid irreversibility in heat exchange. 

Problems 

1. A sheet of steel 10 by 12 ft, yy in. in thickness, separates two flowing gas 
streams which differ in temperature. The surface temperatures of the steel wall 
differ by 2°F. How much heat flows through the wall per minute? 

2. A stationary layer of air fills the space between two parallel steel walls which 
are 2 ft apart. If the interior-surface temperatures of the steel walls are 300 and 
100°F, respectively, calculate the rate of heat transfer per hour per square foot of 
area of each wall. Repeat, changing the surface temperatures to 400 and 200°F. 
Neglect radiation. 

3. A layer of corkboard with an area of 10 ft 2 allows the flow of 90 Btu of heat 
per hour when its surface temperatures are 80 and 20°F. What is its thickness? 
What is its conductance? Its resistance? 

4. A wall has a thickness of 8 in. and an area of 100 ft 2 . When the surface tem¬ 
peratures are 150 and 70°F, the heat transfer is 6500 Btu/hr. What is the conduc¬ 
tivity of the wall? What is its conductance? Its resistance? 

5. A test specimen of brick building wall 8 in. thick allows the transfer of 200 Btu 
during a 30-min test. A temperature differential of 90°F is maintained between the 
two wall surfaces. What is the area of the wall? 

6. A brick-veneer building wall consists of a 4-in. thickness of brick, pine sheathing 
yf in. thick, a 3^-in. space which is filled with mineral wool, and an inner layer of 
wood lath and plaster, y in. in thickness. Calculate (a) the conductance of this wall 
and (6) the temperature at the inner surface of the brick when its outer-surface tem¬ 
perature is 5°F and the temperature of the exposed surface of the plaster is 60°F. 

7. A furnace wall consists of a 9-in. thickness of refractory (fire-clay) brick, a 3-in. 
thickness of insulation [& = 0.5 (Btu)(in.)/(ft 2 )(°F)(hr)j, and 4 in. of common brick. 
The inside-surface temperature of this wall (the refractory-brick surface) is 1400°F, 
and its outside- (common-brick) surface temperature is 110°F. Calculate (a) the 
conductance of the wall and (6) the average temperature of the layer of refractory 
brick. 

8. A steel boiler tube is 3^- in. outside diameter, 0.12 in. thick. Its inside-surface 
temperature is 400°F. If heat is transferred to the water inside the tube at a rate of 
4000 Btu/hr per foot of pipe length, what is the temperature of the outside surface? 


526 


BASIC ENGINEERING THERMODYNAMICS 


9. In Prob. 8 assume that boiler scale (k = 15) has accumulated to a thickness of 
xV in. on the inside of the pipe. The temperature of the inner surface of this scale is 
400°F. Assuming the rate of heat transfer is the same as in Prob. 8, calculate the 
temperature of the outside surface of the pipe. 

10. In Example 20:4(7, assume that the inner surface of the pipe is coated with a 
layer of boiler scale yV in. in thickness (k = 15). The surface temperatures are as in 
the example, except that the surface temperature of 300°F applies to the inner surface 
of the scale. Calculate (a) the percentage reduction in heat transfer and (6) the 
surface temperatures of the steel pipe. 

11. A 3^ in.-outside-diameter steel tube, 10 ft long, carries steam through a passage 
30 in. in diameter and lined with white firebrick. The outside-surface temperature 
of the pipe is 300°F; the inside-surface temperature of the firebrick is 1000°F. Calcu¬ 
late the radiant heat absorbed per hour by the pipe if (a) the pipe is considered to be 
completely enclosed and (6) if the two surfaces are treated as concentric infinite 
cylinders. 

12. A 4-in.-outside-diameter bare steel pipe, 20 ft long, carries steam through a 

room the plastered walls of which have a surface temperature of 100°F. The surface 
temperature of the pipe is 400°F. ( a ) Calculate the heat loss per hour from the pipe 

by radiation. ( b ) By covering the pipe with a 1-in. layer of asbestos insulation, its 
surface temperature may be reduced to 140°F. What is the percentage reduction in 
the heat loss by radiation as compared with part a? 

13. A horizontal bank of water tubes is located 20 ft above and parallel with a 
16- by 22-ft furnace floor. The tubes are staggered so that all radiant heat is inter¬ 
cepted. The side walls of the furnace may reradiate but are assumed to absorb none 
of the radiant heat from the floor, which is constructed of light buff refractory brick 
and has a surface temperature of 2000°F. The surface temperature of the tubes is 
400°F. Calculate the radiant heat received per hour by the tubes. 

14. In Example 20:6A, change the dimensions of the furnace to 18 by 24 ft, with 
a height of 18 ft. The tube bank covers one of the 24-ft vertical walls. Assume the 
tube surface temperature to be 400°F and the surface temperature of the walls to be 
2000°F. Calculate the radiant heat received by the tube bank (a) from each of the 
adjacent vertical walls and (6) from the opposite vertical wall. 

15. In Prob. 1, assume both surface coefficients to be 1.5. What is the total 
difference between the free-stream temperatures of the two gas flows? 

16. In Prob. 2, if the layer of air is not stationary but has a surface coefficient of 2 
at both walls, find the temperature at the center of the layer and the rate of heat 
transfer per hour per square foot of wall area. Assume wall-surface temperatures of 
100 and 300°F, and compare your answer with the answer to Prob. 2. 

17 What temperature gradient in the layers of air immediately adjacent to the 
wall surface corresponds (a) to the surface coefficient of 1.6 suggested for inside 
walls in Art. 20:8? ( b ) To the outside surface coefficient of 6.0? 

18. (a) Calculate the over-all heat-transfer coefficient for the wall described in 
Prob. 6. Assume an outside wind velocity of 15 mph. (6) If the 3-J-in. space had 
not been filled with mineral wool, what would have been the over-all heat-transfer 
coefficient for the wall? 

19. In Prob. 7, assume that the inside-surface coefficient is 4, the outside is 1.6. 
What are the plateau levels of temperature of the hot gases inside and the air outside 
the wall? 

20. Show in detail the steps in the derivation of Eq. (20:15). 

21. Calculate the Prandtl number for air at atmospheric pressure and (a) a tem¬ 
perature of 50°F; (6) a temperature of 200°F. Do the same for hydrogen, and com¬ 
pare. For saturated steam at 212°F. 


THE TRANSMISSION OF HEAT 


527 


22. Calculate the Prandtl number for liquid water at (a) 50°F; (6) at 200°F. 

23. Calculate the average inside film coefficient when 600 ft 3 of air per minute at 
atmospheric pressure and an average temperature of 140°F is carried by a rectangular 
8- by 12-in. duct. 

24. SAE 30 lubricating oil (specific gravity = 0.88; c p = 0.50) is forced through a 
pipe with an internal diameter of 2.07 in. at an average velocity of 10 fps. The aver¬ 
age pressure of the oil is 40 psia, and its average temperature is 150°F. Determine 
the average inside film coefficient. 

25. A 1-in. pipe (internal diameter = 1.05 in.) discharges 30 gal of water per minute 
at 180°F. Determine the inside film coefficient at a point in the pipe where the 
pressure is 70 psia, the temperature 200°F. 

26. Superheated steam at 500 psia, 600°F (see steam tables for its viscosity and 
conductivity), moves through a pipe with an internal diameter of 5.85 in. at a velocity 
of 100 fps. Find the inside film coefficient. 

27. A stream of air is heated in passing through a bank of four staggered rows of 
1-in. steam pipe (1.3 in. outside diameter) in a direction normal to the plane of the 
bank. The outside-surface temperature of the tubes averages 210°F, and the average 
temperature of the air is 80°F. The velocity of air flow is 10 fps in the most con¬ 
stricted section. Calculate the average outside film coefficient. Assume a straight- 
line variation of F with the number of rows of tubes. 

28. In Example 20:4 C, assume that the pipe is horizontal and passes through a 

room with an ambient-air temperature of 70°F. The temperature at the outer 
surface of the insulation is 90°F. (a) Calculate the outside film coefficient of the 

pipe. (6) Based on your answer to part a, calculate the rate of heat flow from the 
outer surface of the pipe per foot of pipe length. Compare with the rate of heat 
transfer through the pipe and insulation, as calculated in the example, and decide 
whether the surface temperature of the insulation is above or below 90°F when the 
ambient-air temperature is 70°F. Determine the surface temperature of the pipe 
for a proper balance of heat flows. 

29. Repeat Prob. 28, but assume the pipe to be vertical. 

30. The condenser tubes of Example 20:11R are made of copper and have a wall 
thickness of 0.08 in. Assume that the water temperature is negligibly lower than 
the outside-surface temperature of the tubes (80°F) and that the average cooling- 
water velocity is 10 fps. (a) Calculate the inside film coefficient. (6) Using the 
outside film coefficient as calculated in the example, compute the over-all coefficient 
of heat transmission (steam to water), (c) Calculate the over-all coefficient of heat 
transmission as based on an outside film coefficient of 1000, and compare with the 
answer to part b. 

31. Steam is condensed at a pressure of 1 in. Hg abs in a surface condenser. The 
tubes are ^in. outside diameter, and their average outside-surface temperature is 60°F. 
Calculate the average outside film coefficient. 

32. If the LMTD between steam and cooling-water temperatures in a surface con¬ 
denser is 20°F, the temperature of the steam is 100°F, and the water enters at 75°F, 
calculate the temperature of the water at exit from the condenser. 

33. In an air preheater, the counterflow principle is employed, and the air enters 
at 70°F and is heated to 250°F. The flue gas enters at 600°F, and its mass rate of 
flow is 10 per cent greater than that of the air. The specific heats of the air and of 
the flue gas may be assumed equal. Calculate the LMTD. 

Symbols 

a absorptivity (= e) 

A area 


528 BASIC ENGINEERING THERMODYNAMICS 

c specific heat 

c p specific heat at constant pressure 
Ci a constant 

C thermal conductance; also, a dimensionless coefficient 
D a dimension; for round pipes, the diameter 
F c configuration factor 
F e emissivity factor 
g acceleration of gravity 

h film, or surface, coefficient in convective heat transfer 

k thermal conductivity; conductance of a homogeneous wall of unit thickness 
L the dimension of length; also, length 
M the dimension of mass; also, mass rate of flow 
Nu Nusselt number ( hD/k ) 

P pressure 

Pr Prandtl number ( c v u/k ) 

Q rate of heat flow 
r radius 

R thermal resistance ( = 1 /C) 

Re Reynolds number 
t scalar temperature 

T the dimension of temperature; also, absolute temperature 
U over-all heat transfer coefficient 
V velocity 
x thickness 

Greek Letters 

/3 coefficient of cubical expansion 
e emissivity 
6 the dimension of time 
n viscosity 
p mass density 
a Stefan’s constant 

Subscripts 

c cold fluid 

cond conducted; by conduction 
/ fluid 
h hot fluid 
i inside 

m mean, or average 
o outside 

p constant pressure 
rad radiated; by radiation 
w wall surface 

0 reference, or base, level of temperature 


APPENDIX 


530 


BASIC ENGINEERING THERMODYNAMICS 


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Keyes. 

Published by John Wiley & Sons, Inc., New York. Reprinted by permission from “Steam, Air and Gas Power,” 3d ed., by Severns and Degler. 











532 


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APPENDIX 


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534 


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APPENDIX 


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538 


BASIC ENGINEERING THERMODYNAMICS 


Table 4. Properties of Mercury Vapor* 
(h and s are measured from 32 F.) 


Pres¬ 

sure 

V 

psia 

Temp 

t, F 

Specific 

volume 

Vg 

ft 3 lb- 1 

Enthalpy, Btu lb" 1 

Entropy, Btu lb m * R 1 

Satu¬ 

rated 

liquid 

h f 

Vapor¬ 

ization 

hfg 

Satu¬ 

rated 

vapor 

hg 

Satu¬ 

rated 

liquid 

Sf 

Vapor¬ 

ization 

s /g 

Satu¬ 

rated 

vapor 

s g 

0.4 

402.3 

114.5 

13.81 

128.1 

141.9 

0.02094 

0.1486 

0.1696 

0.6 

426.1 

78.23 

14.70 

127.6 

142.3 

0.02195 

0.1441 

0.1660 

0.8 

443.8 

59.71 

15.36 

127.2 

142.6 

0.02269 

0.1408 

0.1635 

1.0 

458.1 

48.45 

15.89 

126.9 

142.8 

0.02328 

0.1382 

0.1615 

1.5 

485.1 

33.14 

16.90 

126.3 

143.2 

0.02436 

0.1337 

0.1580 

2 

505.2 

25.31 

17.65 

125.8 

143.5 

0.02514 

0.1304 

0.1556 

3 

535.4 

17.34 

18.78 

125.2 

144.0 

0.02629 

0.1258 

0.1521 

4 

558.0 

13.26 

19.62 

124.7 

144.3 

0.02714 

0.1225 

0.1497 

5 

576.2 

10.77 

20.30 

124.3 

144.6 

0.02780 

0.1200 

0.1478 

6 

591.4 

9.096 

20.87 

123.9 

144.8 

0.02834 

0.1179 

0.1462 

7 

605.0 

7.882 

21.37 

123.6 

145.0 

0.02882 

0.1161 

0.1450 

8 

616.8 

6.963 

21.81 

123.4 

145.2 

0.02923 

0.1146 

0.1439 

9 

627.5 

6.244 

22.21 

123.2 

145.4 

0.02960 

0.1133 

0.1429 

10 

637.3 

5.661 

22.58 

122.9 

145.5 

0.02993 

0.1121 

0.1420 

15 

676.5 

3.892 

24.04 

122.1 

146.1 

0.03124 

0.1074 

0.1387 

20 

706.2 

2.983 

25.15 

121.4 

146.6 

0.03220 

0.1041 

0.1363 

25 

730.4 

2.429 

26.05 

120.9 

146.9 

0.03297 

0.1016 

0.1345 

30 

750.9 

2.053 

26.81 

120.4 

147.2 

0.03360 

0.09953 

0.1331 

35 

769.0 

1.781 

27.49 

120.0 

147.5 

0.03416 

0.09774 

0.1319 

40 

784.8 

1.576 

28.08 

119.7 

147.8 

0.03464 

0.09621 

0.1308 

45 

799.3 

1.414 

28.62 

119.4 

148.0 

0.03507 

0.09486 

0.1299 

50 

812.5 

1.284 

29.11 

119.1 

148.2 

0.03546 

0.09364 

0.1291 

60 

836.1 

1.086 

29.99 

118.6 

148.6 

0.03614 

0.09154 

0.1276 

70 

856.6 

0.9436 

30.75 

118.1 

148.9 

0.03672 

0.08976 

0.1264 

80 

874.8 

0.8349 

31.43 

117.7 

149.1 

0.03725 

0.08824 

0.1254 

90 

891.6 

0.7497 

32.06 

117.3 

149.4 

0.03771 

0.08687 

0.1245 

100 

906.9 

0.6811 

32.63 

117.0 

149.6 

0.03813 

0.08565 

0.1237 

120 

934.4 

0.5767 

33.60 

116.4 

150.1 

0.03887 

0.08353 

0.1224 

140 

958.3 

0.5012 

34.55 

115.9 

150.4 

0.03951 

0.08175 

0.1212 

160 

979.9 

0.4438 

35.35 

115.4 

150.8 

0.04007 

0.08019 

0.1202 

180 

999.6 

0.3990 

36.09 

115.0 

151.1 

0.04058 

0.07881 

0.1193 


♦Reproduced from L. S. Marks, “Mechanical Engineers’ Handbook,” McGraw-Hill Book 
Company, Inc., New York. 



























r 

40" 

38 

36 

34 

32 

30 

28 

26 

24 

22 

20 

18 

16 

14 

12 

10 

8 

6 

4 

2 

0 

2 

4 

6 

8 

10 

12 

14 

16 

18 

20 

22 

24 

26 

28 

30 

32 

34 

36 

38 

40 

42 

44 

46 

48 

60 

52 

54 

56 

58 

60 

62 

64 

66 

68 

70 

72 

74 

76 

78 

80 

82 

84 

86 

88 

90 

92 

94 

96 

98 

00 

05 

10 

15 


APPENDIX 


539 


Table 5. Properties of Saturated Ammonia 1 
(Entropy and enthalpy measurements are from — 40°F.) 


Specific volume, 
cu. ft. per lb. 

Enthalpy 

of 

liquid 

hf 

Enthalpy 

of 

vapori¬ 

zation 

hf 0 

Enthalpy 

of 

vapor 

h a 

Entropy 

of 

liquid 

Sf 

of 

evap. 

Sfg 

of 

vapor 

So 

Liquid 

Vf 

Vapor 

Vq 

0.02322 

24.86 

0.0 

597.6 

597.6 

0.000 

1.4242 

1.4242 

0.02326 

23.53 

2.1 

596.2 

598.3 

0.0051 

1.4142 

1.4193 

0.02331 

22.27 

4.3 

594.8 

599.1 

0.0101 

1.4043 

1.4144 

0.02335 

21.10 

6.4 

593.5 

599.9 

0.0151 

1.3945 

1.4096 

0.02340 

20.00 

8.5 

592.1 

600.6 

0.0201 

1.3847 

1.4048 

0.02345 

18.97 

10.7 

590.7 

601.4 

0.0250 

1.3751 

1.4001 

0.02349 

18.00 

12.8 

589.3 

602.1 

0.0300 

1.3655 

1.3955 

0.02354 

17.09 

14.9 

587.9 

602.8 

0.0350 

1.3559 

1.3909 

0.02359 

16.24 

17.1 

586.5 

603.6 

0.0399 

1.3464 

1.3863 

0.02364 

15.43 

19.2 

585.1 

604.3 

0.0448 

1.3370 

1.3818 

0.02369 

14.68 

21.4 

583.6 

605.0 

0.0497 

1.3277 

1.3774 

0.02374 

13.97 

23.5 

582.2 

605.7 

0.0545 

1.3184 

1.3729 

0.02378 

13.29 

25.6 

580.8 

606.4 

0.0594 

1.3092 

1.3686 

0.02383 

12.66 

27.8 

579.3 

607.1 

0.0642 

1.3001 

1.3643 

0.02384 

12.06 

30.0 

577.8 

607.8 

0.0690 

1.2910 

1.3600 

0.02393 

11.50 

32.1 

576.4 

608.5 

0.0738 

1.2820 

1.3558 

0.02399 

10.97 

34.3 

574.9 

609.2 

0.0786 

1.2730 

1.3516 

0.02404 

10.47 

36.4 

573.4 

609.8 

0.0833 

1.2641 

1.3474 

0.02409 

9.991 

38.6 

571.9 

610.5 

0.0880 

1.2553 

1.3433 

0.02414 

9.541 

40.7 

570.4 

611.1 

0.0928 

1.2465 

1.3393 

0.02419 

9.116 

42.9 

568.9 

611.8 

0.0975 

1.2377 

1.3352 

0.02424 

8.714 

45.1 

567.3 

612.4 

0.1022 

1.2290 

1.3312 

0.02430 

8.333 

47.2 

565.8 

613.0 

0.1069 

1.2204 

1.3273 

0.02435 

7.971 

49.4 

564.2 

613.6 

0.1115 

1.2119 

1.3234 

0.02440 

7.629 

51.6 

562.7 

614.3 

0.1162 

1.2033 

1.3195 

0.02446 

7.304 

53.8 

561.1 

614.9 

0.1208 

1.1949 

1.3157 

0.02451 

6.996 

56.0 

559.5 

615.5 

0.1254 

1.1864 

1.3118 

0.02457 

6.703 

58.2 

557.9 

616.1 

0.1300 

1.1781 

1.3081 

0.02462 

6.425 

60.3 

556.3 

616.6 

0.1346 

1.1697 

1.3043 

0.02468 

6.161 

62.5 

554.7 

617.2 

0.1392 

1.1614 

1.3006 

0.02474 

5.910 

64.7 

553.1 

617.8 

0.1437 

1.1532 

1.2969 

0.02479 

5.671 

66.9 

551.4 

618.3 

0.1483 

1.1450 

1.2933 

0.02485 

5.443 

69.1 

549.8 

618.9 

0.1528 

1.1369 

1.2897 

0.02491 

5.227 

71.3 

548.1 

619.4 

0.1573 

1.1288 

1.2861 

0.02497 

5.021 

73.5 

546.4 

619.9 

0.1618 

1.1207 

1.2825 

0.02503 

4.825 

75.7 

544.8 

620.5 

0.1663 

1.1127 

1.2790 

0.02508 

4.637 

77.9 

543.1 

621.0 

0.1708 

1.1047 

1.2755 

0.02514 

4.459 

80.1 

541.4 

621.5 

0.1753 

1.0968 

1.2721 

0.02521 

4.289 

82.3 

539.7 

622.0 

0.1797 

1.0889 

1.2686 

0.02527 

4.126 

84.6 

537.9 

622.5 

0.1841 

1.0811 

1.2652 

0.02533 

3.971 

86.8 

536.2 

623.0 

0.1885 

1.0733 

1.2618 

0.02539 

3.823 

89.0 

534.4 

623.4 

0.1930 

1.0655 

1.2585 

0.02545 

3.682 

91.2 

532.7 

623.9 

0.1974 

1.0578 

1.2552 

0.02551 

3.547 

93.5 

530.9 

624.4 

0.2018 

1.0501 

1.2519 

0.02558 

3.418 

95.7 

529.1 

624.8 

0.2062 

1.0424 

1.2486 

0.02564 

3.294 

97.9 

527.3 

625.2 

0.2105 

1.0348 

1.2453 

0.02571 

3.176 

100.2 

525.5 

625.7 

0.2149 

1.0272 

1.2421 

0.02577 

3.063 

102.4 

523.7 

626.1 

0.2192 

1.0197 

1.2389 

0.02584 

2.954 

104.7 

521.8 

626.5 

0.2236 

1.0121 

1.2357 

0.02590 

2.851 

106.9 

520.0 

626.9 

0.2279 

1.0046 

1.2325 

0.02597 

2.751 

109.2 

518.1 

627.3 

0.2322 

0.9972 

1.2294 

0.02604 

2.656 

111.5 

516.2 

627.7 

0.2365 

0.9897 

1.2262 

0.02611 

2.565 

113.7 

514.3 

628.0 

0.2408 

0.9823 

1.2231 

0.02618 

2.477 

116.0 

512.4 

628.4 

0.2451 

0.9750 

1.2201 

0.02625 

2.393 

118.3 

510.5 

628.8 

0.2494 

0.9676 

1.2170 

0.02632 

2.312 

120.5 

508.6 

629.1 

0.2537 

0.9603 

1.2140 

0.02639 

2.235 

122.8 

506.6 

629.4 

0.2579 

0.9531 

1.2110 

0.02646 

2.161 

125.1 

504.7 

629.8 

0.2622 

0.9458 

1.2080 

0.02653 

2.089 

127.4 

502.7 

630.1 

0.2664 

.0.9386 

1.2050 

0.02661 

2.021 

129.7 

500.7 

630.4 

0.2706 

0.9314 

1.2020 

0.02668 

1.955 

132.0 

498.7 

630.7 

0.2749 

0.9242 

1.1991 

0.02675 

1.892 

134.3 

496.7 

631.0 

0.2791 

0.9171 

1.1962 

0.02684 

1.831 

136.6 

494.7 

631.3 

0.2833 

0.9100 

1.1933 

0.02691 

1.772 

138.9 

492.6 

631.5 

0.2875 

0.9029 

1.1904 

0.02699 

1.716 

141.2 

490.6 

631.8 

0.2917 

0.8958 

1.1875 

0.02707 

1.661 

143.5 

488.5 

632.0 

0.2958 

0.8888 

1.1846 

0.02715 

1.609 

145.8 

486.4 

632.2 

0.3000 

0.8818 

1.1818 

0.02723 

1.559 

148.2 

484.3 

632.5 

0.3041 

0.8748 

1.1789 

0.02731 

1.510 

150.5 

482.1 

632.6 

0.3083 

0.8678 

1.1761 

0.02739 

1.464 

152.9 

480.0 

632.9 

0.3125 

0.8608 

1.1733 

0.02747 

1.419 

155.2 

477.8 

633.0 

0.3166 

0.8539 

1.1705 

0.02769 

1.313 

161.1 

472.3 

633.4 

0.3269 

0.8366 

1.1635 

0.02790 

1.217 

167.0 

466.7 

633.7 

0.3372 

0.8194 

1.1566 

0.02813 

1.128 

173.0 

460.9 

633.9 

0.3474 

0.8023 

1.1497 

0.02836 

1.047 

179.0 

455.0 

634.0 

0.3576 

0.7851 

1.1427 


Engineers’ Handbook.” 


































540 


BASIC ENGINEERING THERMODYNAMICS 


Table 6. Properties of Superheated Ammonia 1 
(Condensed from Circular No. 142 of the U. S. Bureau of Standards, 1923) 
v = specific volume in cu. ft. per lb.; h = enthalpy in B.t.u. per lb.; 

s = entropy. 

(h and s are measured from — 40°F.) 


Pressure 
lb. per 

Temp, 
of satu¬ 
rated 


Temperature of superheated vapor, °F. 

sq. in. 












abs. 

Vapor, 

°F. 


-30 

-20 

O 

r-H 

1 

0 

10 

20 

30 

40 

50 



V 

26.58 

27.26 

27.92 

28.58 

29.24 

29.90 

30.55 

31.20 

31.85 

10 

-41.34 

h 

603.2 

608.5 

613.7 

618.9 

624.0 

629.1 

634.2 

639.3 

644.4 



s 

1.4420 

1.4542 

1.4659 

1.4773 

1.4884 

1.4992 

1.5097 

1.5200 

1.5301 



V 



13.74 

14.09 

14.44 

14.78 

15.11 

15.45 

15.78 

20 

—16.64 

h 



610.0 

615.5 

621.0 

626.4 

631.7 

637.0 

642.3 



s 



1.3784 

1.3907 

1.4025 

1.4138 

1.4240 

1.4356 

1.4460 



V 




9.250 

9.492 

9.731 

9.966 

10.20 

10.43 

30 

- 0.57 

h 




611.9 

617.8 

623.5 

629.1 

634.6 

640.1 



s 




1.3371 

1.3497 

1.3618 

1.3733 

1.3845 

1.3953 



v 






7.203 

7.387 

7.568 

7.746 

40 

11.66 

h 






620.4 

626.3 

632.1 

637.8 



s 






1.3231 

1.3353 

1.3470 

1.3583 



x 







5.838 

5.988 

6.135 

50 

21.67 

h 







623.4 

629.5 

635.4 



s 







1.3046 

1.3169 

1.3286 
















100 

12C 

140 

160 

180 

200 

240 

280 

320 



V 

4.190 

4.371 

4.548 

4.722 

4.893 

5.063 

5.398 

5.730 


80 

44.40 

h 

658.7 

670.4 

681.8 

693.2 

704.4 

715.6 

738.1 

760.7 




s 

1.3199 

1.3404 

1.3598 

1.3784 

1.3963 

1.4136 

1.4467 

1.4781 

. 



V 

3.304 

3.454 

3.600 

3.743 

3.883 

4.021 

4.294 

4.562 


100 

56.05 

h 

655.2 

667.3 

679.2 

690.8 

702.3 

713.7 

736.5 

759.4 




s 

1.2891 

1.3104 

1.3305 

1.3495 

1.3678 

1.3854 

1.4190 

1.4507 

. 



V 

2.712 

2.842 

2.967 

3.089 

3.209 

3.326 

3.557 

3.783 


120 

66.02 

h 

651.6 

664.2 

676.5 

688.5 

700.2 

711.8 

734.9 

758.0 




s 

1.2628 

1.2850 

1.3058 

1.3254 

1.3441 

1.3620 

1.3960 

1.4281 

. 



V 

2.288 

2.404 

2.515 

2.622 

2.727 

2.830 

3.030 

3.227 

3.420 

140 

74.79 

h 

647.8 

661.1 

673.7 

686.0 

698.0 

709.9 

733.3 

756.7 

780.0 



s 

1.2396 

1.2628 

1.2843 

1.3045 

1.3236 

1.3418 

1.3763 

1.4088 

1.4395 



V 

1.969 

2.075 

2.175 

2.272 

2.365 

2.457 

2.635 

2.809 

2.980 

160 

82.64 

h 

643.9 

657.8 

670.9 

683.5 

695.8 

707.9 

731.7 

755.3 

778.9 



s 

1.2186 

1.2429 

1.2652 

1.2859 

1.3054 

1.3240 

1.3591 

1.3919 

1.4229 



V 

1.720 

1.818 

1.910 

1.999 

2.084 

2.167 

2.328 

2.484 

2.637 

180 

89.78 

h 

639.9 

654.4 

668.0 

681.0 

693.6 

705.9 

730.1 

753.9 

777.7 



s 

1.1992 

1.2247 

1.2477 

1.2691 

1.2891 

1.3081 

1.3436 

1.3768 

1.4081 



V 

1.520 

1.612 

1.698 

1.780 

1.859 

1.935 

2.082 

2.225 

2.364 

200 

96.34 

h 

635.6 

650.9 

665.0 

678.4 

691.3 

703.9 

728.4 

752.5 

776.5 



s 

1.1809 

1.2077 

1.2317 

1.2537 

1.2742 

1.2935 

1.3296 

1.3631 

1.3947 



V 


1.443 

1.525 

1.601 

1.675 

1.745 

1.881 

2.012 

2.140 

220 

102.42 

h 


647.3 

662.0 

675.8 

689.1 

701.9 

726.8 

751.1 

775.3 



s 


1.1917 

1.2167 

1.2394 

1.2604 

1.2801 

1.3168 

1.3507 

1.3825 





1.302 

1.380 

1.452 

1.521 

1.587 

1.714 

1.835 

1.954 

240 

108.09 

h 


643.5 

658.8 

673.1 

686.7 

699.8 

725.1 

749.8 

774.1 



s 


1.1764 

1.2025 

1.2259 

1.2475 

1.2677 

1.3049 

1.3392 

1.3712 



V 


1.182 

1.257 

1.326 

1.391 

1.453 

1.572 

1.686 

1.796 

260 

113.42 

h 


639.5 

655.6 

670.4 

684.4 

697.7 

723.4 

748.4 

772.9 



s 


1.1617 

1.1889 

1.2132 

1.2354 

1.2560 

1.2938 

1.3285 

1.3608 


1 Marks, “Mechanical Engineers’ Handbook.” 








































































APPENDIX 


541 


Table 7., Properties of Saturated Sulphur Dioxide 1 
(h and s are measured from — 40°F.) 


Temp., 

op 

Pres¬ 
sure, lb. 
per 

Specific 
cu. ft. 

volume, 
per lb. 

En¬ 

thalpy 

of 

liquid, 
B.t.u. 
per lb. 

En¬ 

thalpy 

of 

vapori¬ 

zation, 

B.t.u. 

En¬ 

thalpy 

of 

Entropy 


sq. in. 
abs. 

Liquid 

Vapor 

vapor, 

B.t.u. 

Of 

liquid 

Of vapor¬ 
ization 

Of vapor 

t 

V 

Vf 

Vg 

hf 

hfg 

hg 

91 

8fg 

8g 

-40 

3.136 

0.01044 

22.42 

0.00 

178.6 

178.6 

0.0000 

0.4256 

0.4256 

-30 

4.331 

0.01053 

16.56 

2.93 

177.0 

179.9 

0.00674 

0.4119 

0.4186 

-20 

5.883 

0.01062 

12.42 

5.98 

175.1 

181.1 

0.01366 

0.3983 

0.4119 

-10 

7.863 

0.01072 

9.44 

9.16 

173.0 

182.1 

0.02075 

0.3847 

0.4054 

0 

10.35 

0.01082 

7.28 

12.44 

170.6 

183.1 

0.02795 

0.3712 

0.3992 

10 

13.42 

0.01092 

5.682 

15.80 

168.1 

183.9 

0.03519 

0.3570 

0.3931 

20 

17.18 

0.01103 

4.487 

19.20 

165.3 

184.5 

0.04241 

0.3447 

0.3871 

30 

21.70 

0.01114 

3.581 

22.64 

162.4 

185.0 

0.04956 

0.3316 

0.3812 

40 

27.10 

0.01126 

2.887 

26.12 

159.3 

185.4 

0.05668 

0.3187 

0.3754 

50 

33.45 

0.01138 

2.348 

29.61 

156.0 

185.6 

0.06370 

0.3060 

0.3697 

60 

40.93 

0.01150 

1.926 

33.10 

152.5 

185.6 

0.07060 

0.2935 

0.3641 

70 

49.62 

0.01163 

1.590 

36.58 

148.9 

185.5 

0.07736 

0.2811 

0.3585 

80 

59.68 

0.01176 

1.321 

40.05 

145.1 

185.2 

0.08399 

0.2690 

0.3529 

90 

71.25 

0.01190 

1.104 

43.50 

141.2 

184.7 

0.09038 

0.2569 

0.3473 

100 

84.52 

0.01204 

0.9262 

46.90 

137.2 

184.1 

0.09657 

0.2452 

0.3417 

110 

99.76 

0.01219 

0.7804 

50.26 

133.1 

183.3 

0.1025 

0.2336 

0.3361 

120 

120.9 

0.01236 

0.6598 

53.58 

128.8 

182.4 

0.1083 

0.2222 

0.3305 

130 

136.5 

0.01253 

0.5595 

56.85 

124.4 

181.2 

0.1138 

0.2110 

0.3247 

140 

158.6 

0.01272 

0.4758 

60.04 

119.9 

179.9 

0.1189 

0.1999 

0.3189 


1 Marks, “Mechanical Engineers’ Handbook.” 

























542 BASIC ENGINEERING THERMODYNAMICS 


Table 8. Properties of Superheated Sulphur Dioxide 1 
(w = specific volume in cu. ft. per lb.; h = enthalpy in B.t.u. per lb.; 

$ = entropy) 

(h and s are measured from — 40°F.) 


Pressure 
lb. per 

Temp, 
of satu¬ 
rated 


Temperature of superheated vapor, °F. 

sq. in. 












aha. 

vapor, 












°F. 


0 

20 

40 

60 

80 

100 

120 

140 

160 



V 

12.75 

13.34 

13.93 

14.52 

15.11 

15.69 

16.26 

16.82 

17.35 

6 

-19.37 

h 

184.3 

187.5 

190.7 

193.9 

197.2 

200.5 

203.8 

207.1 

210.4 



s 

0.4185 

0.4254 

0.4320 

0.4383 

0.4444 

0.4504 

0.4561 

0.4618 

0.4672 



V 

7.545 

7.939 

8.316 

8.681 

9.038 

9.389 

9.736 

10.08 

10.42 

10 

- 1.34 

h 

183.2 

186.7 

190.1 

193.5 

196.9 

200.3 

203.7 

207.1 

210.5 



s 

0.4005 

0.4080 

0.4151 

0.4216 

0.4280 

0.4341 

0.4400 

0.4457 

0.4512 



V 


5.192 

5.470 

5.734 

5.988 

6.233 

6.471 

6.705 

6.937 

15 

14.43 

h 


185.4 

189.2 

192.8 

196.4 

199.9 

203.3 

206.7 

210.1 



s 


0.3927 

0.4005 

0.4078 

0.4144 

0.4208 

0.4268 

0.4326 

0.4383 



V 



4.035 

4.251 

4.454 

4.648 

4.834 

5 015 

5.193 

20 

26.44 

h 



187.8 

191.8 

195.6 

199.3 

202.9 

206 5 

209.9 



s 



0.3896 

0.3972 

0.4043 

0.4109 

0.4173 

0 4232 

0 4290 



V 



3.181 

3.363 

3.536 

3.696 

3 848 

3 998 

4 145 

25 

36.33 

h 



186.1 

190.6 

194.7 

198 6 

202 4 

206 0 

209 6 



s 



0.3793 

0.3880 

0.3958 

0.4029 

0.4095 

0.4157 

0.4216 




60 

80 

100 

120 

140 

160 

180 

200 

220 



V 

2.747 

2.907 

3.052 

3.189 

3.318 

3.443 

3.565 

3.685 

3.803 

30 

44.76 

h 

189.3 

193.8 

197.9 

201.8 

205.6 

209.3 

212.9 

216.5 

220.1 



s 

0.3797 

0.3885 

0.3960 

0.4029 

0.4094 

0.4154 

0.4211 

0.4266 

0.4318 



V 

1.980 

2.121 

2.246 

2.360 

2.465 

2.565 

2.662 

2.755 

2.845 

40 

58.83 

h 

185.9 

191.3 

196.1 

200.4 

204.6 

208.5 

212.3 

216.0 

291.7 



s 

0.3654 

0.3754 

0.3842 

0.3918 

0.3988 

0.4053 

0.4113 

0.4169 

0.4223 



V 



1.288 

1.403 

1.514 

1 608 

1 689 

1 751 

1 819 

60 

80.29 

h 



191.4 

197.0 

201 9 

206 5 

210 7 

214 8 

218 7 



s 



0.3640 

0 3738 

0 3822 

0 3896 

0 3964 

0 4026 

0 4084 



V 



0.993 

1.084 

1 163 

1 232 

1 292 

1 347 

1 400 

80 

96.88 

h 



185.6 

192 5 

198 6 

203 9 

208 7 

213 3 

217 5 



s 



0.3457 

0.3580 

0.3682 

0.3769 

0.3846 

0.3915 

0.3978 


1 Marks, “Mechanical Engineers’ Handbook.” 

























































APPENDIX 


543 


Table 9. Properties of Carbon Dioxide 1 


(Enthalpy measurements are from 32°F.) 


Tem¬ 

pera¬ 

ture, 

°F. 

Pres- 

Density, 
lb. per cu. ft. 

B.t.u. per lb. 

Entropy 

sure, lb. 
per 
sq. in. 
abs. 

of the 
liquid 

1 

of the 

vapor 

Enthalpy 
of the 
liquid 

Enthalpy 
of vapor¬ 
ization 

Enthalpy 
of satu¬ 
rated 

vapor 

of the 
liquid 

of the 

vapor 

t 

P 

1 

Vf 

I 

Vg 

hf 

hfo 

Jig 

Sf 

S 0 

-40 

145.87 

69.8 

1.64 

-38.5 

136.5 

98.0 

-0.0850 

0.2400 

-35 

161.33 

69.1 

1.83 

-35.8 

134.3 

98.5 

-0.0793 

0.2367 

-30 

177.97 

68.3 

2.02 

-33.1 

132.1 

99.0 

-0.0735 

0.2336 

-25 

195.85 

67.6 

2.23 

-30.4 

129.8 

99.4 

-0.0676 

0.2306 

-20 

215.02 

66.9 

2.44 

-27.7 

127.5 

99.8 

-0.0619 

0.2277 

-15 

235.53 

66.1 

2.66 

-24.9 

125.0 

100.1 

-0.0560 

0.2250 

-10 

257.46 

65.3 

2.91 

-22.1 

122.4 

100.3 

-0.0500 

0.2220 

- 5 

280.85 

64.5 

3.17 

-19.4 

120.0 

100.6 

-0.0440 

0.2198 

0 

305.76 

63.6 

3.46 

-16.7 

117.5 

100.8 

-0.0381 

0.2173 

5 

332.2 

62.8 

3.77 

-14.0 

115.0 

101.0 

-0.0322 

0.2151 

10 

360.4 

61.9 

4.12 

-11.2 

112.2 

101.0 

-0.0264 

0.2124 

15 

390.2 

61.0 

4.49 

- 8.4 

109.4 

101.0 

-0.0204 

0.2100 

20 

421.8 

60.0 

4.89 

- 5.5 

106.3 

100.8 

-0.0144 

0.2071 

25 

455.3 

59.0 

5.33 

- 2.5 

103.1 

100.6 

-0.0083 

0.2043 

30 

490.6 

58.0 

5.81 

+ 0.4 

99.7 

100.1 

-0.0021 

0.2012 

35 

528.0 

57.0 

6.35 

3.5 

95.8 

99.3 

+0.0039 

0.1975 

40 

567.3 

55.9 

6.91 

6.6 

91.8 

98.4 

0.0099 

0.1934 

45 

608.9 

54.7 

7.60 

9.8 

87.5 

97.3 

0.0160 

0.1892 

50 

652.7 

53.4 

8.37 

12.9 

83.2 

96.1 

0.0220 

0.1852 

55 

698.8 

52.1 

9.27 

16.1 

78.7 

94.8 

0.0282 

0.1809 

60 

747.4 

50.7 

10.2 

19.4 

74.0 

93.4 

0.0345 

0.1767 

65 

798.6 

49.1 

11.3 

22.9 

68.9 

91.8 

0.0412 

0.1724 

70 

852.4 

47.3 

12.6 

26.6 

62.7 

89.3 

0.0482 

0.1665 

75 

909.3 

45.1 

14.2 

30.9 

54.8 

85.7 

0.0562 

0.1587 

80 

969.3 

42.4 

16.2 

35.6 

44.0 

79.6 

0.0649 

0.1464 

85 

1032.7 

38.2 

19.1 

41.7 

27.5 

69.2 

0.0761 

0.1265 

88 

1072.1 

32.9 

25.4 

Critical Point 





1 Marks, “Mechanical Engineers’ Handbook.’* 































544 BASIC ENGINEERING THERMODYNAMICS 


Table 10. Properties of Dichlorodifluoromethane (F-12-Freon) 1 
(Entropy and enthalpy measurements are from — 40°F.) 


Temp., °F. 
t 

Pressure, 
p.s.i. abs. 

V 

Specific volume, 
cu. ft. per lb. 

Enthalpy 

Entropy 

Liquid 

Vf 

Vapor 

Vg 

Liquid 

hf 

Vapor 

Jig 

Liquid 

Sf 

Vapor 

Sg 

-40 

9.33 

0.0106 

3.911 

0 

73.50 

0 

0.17517 

— 30 

12.02 

0.0107 

3.088 

2.03 

74.70 

0.00471 

0.17387 

— 20 

15.28 

0.0108 

2.474 

4.07 

75.87 

0.00940 

0.17275 

— 10 

19.20 

0.0109 

2.003 

6.14 

77.05 

0.01403 

0.17175 

0 

23.87 

0.0110 

1.637 

8.25 

78.21 

0.01869 

0.17091 

2 

24.89 

0.0111 

1.574 

8.67 

78.44 

0.01961 

0.17075 

4 

25.96 

0.0111 

1.514 

9.10 

78.67 

0.02052 

0.17060 

6 

27.05 

0.0111 

1.457 

9.53 

78.90 

0.02143 

0.17045 

8 

28.18 

0.0111 

1.403 

9.96 

79.13 

0.02235 

0.17030 

10 

29.35 

0.0112 

1.351 

10.39 

79.36 

0.02328 

0.17015 

12 

30.56 

0.0112 

1.301 

10.82 

79.59 

0.02419 

0.17001 

14 

31.80 

0.0112 

1.253 

11.26 

79.82 

0.02510 

0.16987 

16 

33.08 

0.0113 

1.207 

11.70 

80.05 

0.02601 

0.16974 

18 

34.40 

0.0113 

1.163 

12.12 

80.27 

0.02692 

0.16961 

20 

35.75 

0.0113 

1.121 

12.55 

80.49 

0.02783 

0.16949 

22 

37.15 

0.0113 

1.081 

13.00 

80.72 

0.02873 

0.16938 

24 

38.58 

0.0114 

1.043 

13.44 

80.95 

0.02963 

0.16926 

26 

40.07 

0.0114 

1.007 

13.88 

81.17 

0.03053 

0.16913 

28 

41.59 

0.0114 

0.973 

14.32 

81.39 

0.03143 

0.16900 

30 

43.16 

0.0115 

0.939 

14.76 

81.61 

0.03233 

0.16887 

32 

44.77 

0.0115 

0.908 

15.21 

81.83 

0.03323 

0.16876 

34 

46.42 

0.0115 

0.877 

15.65 

82.05 

0.03413 

0.16865 

36 

48.13 

0.0115 

0.848 

16.10 

82.27 

0.03502 

0.16854 

38 

49.88 

0.0116 

0.819 

16.55 

82.49 

0.03591 

0.16843 

40 

51.68 

0.0116 

0.792 

17.00 

82.71 

0.03680 

0.16833 

42 

53.51 

0.0116 

0.767 

17.46 

82.93 

0.03770 

0.16823 

44 

55.40 

0.0117 

0.742 

17.91 

83.15 

0.03859 

0.16813 

46 

57.35 

0.0117 

0.718 

18.36 

83.36 

0.03948 

0.16803 

48 

59.35 

0.0117 

0.695 

18.82 

83.57 

0.04037 

0.16794 

50 

61.39 

0.0118 

0.673 

19.27 

83.78 

0.04126 

0.16785 

52 

63.49 

0.0118 

0.652 

19.72 

83.99 

0.04215 

0.16776 

54 

65.63 

0.0118 

0.632 

20.18 

84.20 

0.04304 

0.16767 

56 

67.84 

0.0119 

0.612 

20.64 

84.41 

0.04392 

0.16758 

58 

70.10 

0.0119 

0.593 

21.11 

84.62 

0.04480 

0.16749 

60 

72.41 

0.0119 

0.575 

21.57 

84.82 

0.04568 

0.16741 

62 

74.77 

0.0120 

0.557 

22.03 

85.02 

0.04657 

0.16733 

64 

77.20 

0.0120 

0.540 

22.49 

85.22 

0.04745 

0.16725 

66 

79.67 

0.0120 

0.524 

22.95 

85.42 

0.04833 

0.16717 

68 

82.24 

0.0121 

0.508 

23.42 

85.62 

0.04921 

0.16709 

70 

84.82 

0.0121 

0.493 

23.90 

85.82 

0.05009 

0.16701 

72 

87.50 

0.0121 

0.479 

24.37 

86.02 

0.05097 

0.16693 

74 

90.20 

0.0122 

0.464 

24.84 

86.22 

0.05185 

0.16685 

76 

93.00 

0.0122 

0.451 

25.32 

86.42 

0.05272 

0.16677 

78 

95.85 

0.0122 

0.438 

25.80 

86.61 

0.05359 

0.16669 

80 

98.76 

0.0123 

0.425 

26.28 

86.80 

0.05446 

0.16662 

82 

101.7 

0.0123 

0.413 

26.76 

86.99 

0.05534 

0.16655 

84 

104.8 

0.0124 

0.401 

27.24 

87.18 

0.05621 

0.16648 

86 

107.9 

0.0124 

0.389 

27.72 

87.37 

0.05708 

0.16640 

88 

111.1 

0.0124 

0.378 

28.21 

87.56 

0.05795 

0.16632 

90 

114.3 

0.0125 

0.368 

28.79 

87.74 

0.05882 

0.16624 

92 

117.7 

0.0125 

0.357 

29.19 

87.92 

0.05969 

0.16616 

94 

121.0 

0.0125 

0.347 

29.68 

88.10 

0.06056 

0.16608 

96 

124.5 

0.0126 

0.338 

30.18 

88.28 

0.06143 

0.16600 

98 

128.0 

0.0126 

0.328 

30.67 

88.45 

0.06230 

0.16592 

100 

131.6 

0.0127 

0.319 

31.16 

88.62 

0.06316 

0.16584 

110 

150.7 

0.0129 

0.277 

33.65 

89.43 

0.06749 

0.16542 

120 

171.8 

0.0131 

0.240 

36.16 

90.15 

0.07180 

0.16495 


1 Adapted from “Refrigerating Data Book.” 





























Temperature - Degrees Fahrenheit 


1000 


appendix 

Entropy 
1.0 


000 





Temperature - Degrees Fahrenheit 
















































































































APPENDIX 



a 

a 


ENTROPY 

Reprinted by permission from “Steam, Air and Gas Power,” by Severn and Degler. Published by John Wiley and Sons, 
















































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































— 250-q- 


r- 240-5 

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APPENDIX 



Dry Bulb Temperature—Degrees F. 

PLATH 3. Psychrometric chart. (Copyright 1942, General Electric Company; reproduced by special permission.) 


Weight of Water Vapor in One Pound of Dry Air—Crain* 





































































































































































































































































































































































































































































































































































































































INDEX 


A 

Absolute humidity, 227 
Absolute pressure, 8-9 
Absolute temperature, 10-11, 83-86 
Absolute zero, of entropy, 94 
of pressure, 8 

of temperature, 10-11, 83-86 
Absorber, 372 

Absorption system of refrigeration, 370- 
375, 383-384 

Absorptivity, 502-503, 505 
Adiabatic process, 15, 33, 207-211 
mixing, 207-211 
reversible, 61-63, 97, 184, 437 
Adiabatic saturation, 217 
Admission, 306, 315 
Adsorption, 221 

Adsorption refrigerating machine, 375- 
376 

Air, 169, 173 

atmospheric (see Atmospheric air) 
in combustion, 460, 464 
excess, 293, 460, 465 
free, 195 

at low pressures (table), 439 
theoretical, for combustion, 465 
Air compressor, axial-flow, 263-264 
piston, 188-196 
rotary, 188, 191-192 
Air-fuel ratio, 465 
Air preheaters, 461, 487 
Air-standard cycle, 269-270 
American Society, of Heating and Venti¬ 
lating Engineers, 495n. 
of Mechanical Engineers, 519n. 
of Refrigerating Engineers, 495n. 
Ammonia, 370-375 

Ammonia-absorption refrigerating sys¬ 
tem, 370-375, 383-384 
Ammonia-compression refrigerating sys¬ 
tem, 367-368 
Ammonia tables, 539-540 
Ammonia-water mixtures, 370-371 


Analysis, of ash, 469 
dimensional, 411-415, 513-514 
flue-gas, 468 

of gas mixtures, gravimetric, 205 
Orsat, 205, 467-469 
volumetric, 205 
of products, 467-469 
simplified, of Otto-cycle engine, 480 
steady-flow, of reciprocating engine, 
55-56, 311 

ultimate, of fuels, 466 
Anticatalyst, 483 
Aqua ammonia, 370-371 
Arithmetical mean temperature differ¬ 
ence (AMTD), 524 
Athodyd, 296 
Atmosphere, standard, 8 
Atmospheric air, 212-224 
absolute humidity of, 227 
adiabatic saturation of, 217 
dew point of, 216 
engineering processes for, 221-224 
properties of, 212-221 
relative humidity of, 215-217 
saturated, 213-214 
specific humidity of, 215-216 
Atmospheric pressure, 8 
Atomic balance in combustion, 464 
Availability, change of, 390 
concept of, 108 
and reversibility, 388 
in steady flow, 395-396 
of system, 387-392 
Availability function, 390 
Available energy, 108-109 
Avogadro’s law, 172 
Axial thrust, 259 

B 

b function, 395 
Beardsley, M. W., 178n., 434 
Beattie, J. W., 449, 454 
Beattie-Bridgman equation, 449-450, 454 
constants for (table), 449 


545 


546 


BASIC ENGINEERING THERMODYNAMICS 


Beau de Rochas, 270 
Bernoulli equation, 49-51 
Berthelot equation, 448-449 
Beta function, 391 
Binary-vapor cycle, 350-354 
Black body, 502, 503 
Blading, of axial-flow compressor, 263 
erosion of, 259 
flow through, 238 
impulse, 238, 251, 252, 254 
reaction, 256, 257 
Blasius equation, 417 
Blower, 188, 191-192 
Boiler, 54, 328, 330, 337, 341, 343 
Bomb calorimeter, 470-471 
Boyle’s law, 168, 208 
Bray ton cycle, 285-291 
Bridgman, P. W., 449 
British thermal unit (Btu), 21-22 
Bucket, entry and exit angles of, 250 
as part of turbine, 230 
velocity of, 250-255 

Butane, combustion analysis for, 464- 
466, 468-469 

C 

Calorimeter, bomb, 470-471 
gas, 472 
separating, 142 
throttling, 142-145 
Capacity, refrigerating, 359 
Capillarity, 12 
Carbon, as fuel, 458 
in refuse, 469 
Carbon dioxide, 173, 370 
absorption of radiant heat by, 508 
as combustion product, 458 
Carbon dioxide tables, 543 
Carbon monoxide, 173 
as combustion product, 458 
Carnot, Sadi, 69, 81, 83 
Carnot cycle, 69 

effect of friction on, 73 
efficiency of, 73, 85, 104 
as refrigeration cycle, 359-364 
for saturated-vapor systems, 327-329 
as used for calculation of constructive 
work, 388 

Carnot engine, 69-72, 268, 297-298 
Carnot heat pump, 71 


Carnot principle, 81-83 
corollaries of, 82 

Carnot refrigerating machine, 359-364 
Centigrade temperature scale, 10 
Centipoise, 410 
Charles’ law, 168 
Chart, Mollier, 101-102, 140-142 
psychrometric, 218-221 
Chemical energy, 7, 471n. 

Chemical equation, 6, 462-467 
Chemical equilibrium, 478 
Clapeyron relation, 163-164 
Clausius, Rudolf, 78 

Clearance, 193-196, 275-276, 280-281, 
308-311 

Clearance factor, 193-196 
Closed system, 1-2 
and First Law, 21-38 
Closed-system process, 22 
and reversibility, 60-61 
Coal, combustion analyses of, 466-467, 
469-471 

heat of combustion of, 475-476 
Coefficient, of cubical expansion, 493, 518 
of discharge, 241-244 
film, 509-511 
of friction, 414, 418 
over-all of heat transmission, 511-513 
of performance, heat-pump, 380 
refrigeration, 360-361, 374 
of velocity, 241-244 
Cold body, 358 
Cold room, 363 
Combustion, 6, 458 
air in, 460, 464 
atomic balance in, 464 
equation of, 462-467 
external, and closed cycle, 297-301 
vs. internal, 334 
heat of, 459, 470-476 

constant-pressure, 472, 473 
vs. constant-volume, 474-476 
constant-volume, 471 
internal, 268 
mass balance in, 462 
reversible, 483 
as source, of heat, 6 
of power, 459, 482-484 
thermodynamics of, 458-486 
volumetric relations in, 463 
Combustion equation, 462-467 


INDEX 


547 


Compressed liquid, 133-135 
Compressibility factor, 445-446, 448 
451-453 

Compression of gases, continuous, 188- 
192 

multistage, 192-193 
in reciprocating-engine cycle, 306 
Compression ratio, 275 
Compression refrigerating machine, 364- 
368 

Compressors, axial-flow, 263-264 
gas, 188-191 
vapor, 328, 363 

Concentration of aqua-ammonia solu¬ 
tions, 370-371 
Condensation, of air, 376ft. 
cylinder, 315, 319-321, 323 
drop, 520 

following metastable expansion, 247 
initial, 314 
nucleus of, 247 

of vapor from mixture, 212-215 
Condenser, 54, 328, 330, 337, 341, 343 
jet, 487 
surface, 487 

Condition line, 242, 261-264, 402 
Conductance, 496 

Conduction of heat, 488-489, 493-501, 
509-513 

Conductivity, thermal, 137, 493-496 
Configuration factor, 504-506 
Conservation of energy, law of, 1 
Conservative system, 49-51 
Constant, equilibrium, 478 
perfect-gas, 171-172 
universal gas, 172 

Constant-pressure heat of combustion 
{see Combustion, heat of) 
Constant-pressure process, 29, 64-65, 
97-99, 184, 436 

Constant-temperature coefficient, 159 
Constant-volume heat of combustion, 471 
Constant-volume process, 27, 65-67, 97- 
99, 184, 436 

Continuity equation of steady flow, 52-53 
Convection, 490-493, 509-521 
Converging-diverging nozzle, 236-237 
Converging nozzle, 232, 236 
Cooling coils, 299, 373 
Cooling tower, 222-223 
Corliss valves, 320 


Corresponding states, law of, 450-451 
Counterflow, 321, 524 
Criterion of stability, 106 
Critical constants, 445 
Critical point, 119, 127-128, 369, 444-445 
Critical pressure, 119 
Critical-pressure ratio, 235, 239-240 
Critical tables, international, 445ft, 495ft. 
Croft, H. O., 504ft., 505ft., 506n., 509ft. 
Cubical expansion, coefficient of, 493, 
518 

Cushion steam, 308 
Cutoff, 279-280, 306 
Cutoff governing, 258, 321 
Cycle, 15 

air-standard, 269-270 
Brayton, 285-291 
Carnot, 69, 273 

closed, for gas-turbine power plant, 
300-301 

closed-system, 36-37 
diesel, 277-281 
dual, 281-283 
Ericsson, 298 
four- and two-stroke, 270 
Holzwarth, 297 
Lenoir, 285 
machine, 269 
open-system, 54 
real-engine, 310-311 
regenerative, 298-300 
reheat, 336-339 
reversible, 69-74 
Stirling, 73, 273 
Cycles, comparison of, 283-284 
Cylinder condensation, 315, 323 
methods of reducing, 319-321 

D 

Dalton’s law, 171, 203 
Dead center, 306 
Dead state, 112-113, 387-392 
Degrees of freedom of motion, 178-179 
Dehumidification of moist air, 217, 221, 
222 

Density, 11 

Dependent property, 16 
Dew point, 215, 216 
Diagram, indicator, 307-311, 315, 324 
phase, 124-125 


548 


BASIC ENGINEERING THERMODYNAMICS 


Diagram, pressure-volume ( see Pressure- 
volume diagram) 
psychrometric, 218-221 
temperature-entropy (see Tempera¬ 
ture-entropy diagram) 

Diagram factor, 324 
Diesel, Dr. Rudolf, 277 
Diesel cycle, 277-281 
Diesel engine, 277-278 
Dieterici equation, 448 
Diffuser, 232 

Dimensional analysis, as applied, to flow, 
411-415 

to heat transfer, 513-514, 518-519 
limitations of, 414 

Dimensions, engineering quantities ex¬ 
pressed in terms of, 411-412, 514 
Discharge, coefficient of, 241, 243-244 
Displacement, piston, 194-196, 275, 276, 
310, 317 

Dissociation, 460, 477-480 
Drag, skin-friction, 430 
viscous, 409-411 
Drop condensation, 520 
Dry-bulb temperature, 215, 217 
Dry compression, 365-366 
Dual cycle, 281-283 
Ducts, flow through, 408-432 

E 

e, definition of, 13 
vs. u, 13, 45 
Eberhardt, J. E., 480 
Economizer, 461, 487 
Effectiveness, 401 

of Carnot refrigeration cycle, 360 
vs. efficiency of turbine stage, 400-403 
Efficiency, adiabatic, 264 

effect of rotative speed on, 253 
vs. effectiveness of turbine stage, 400- 
403 

engine, 312-313 
heat, 38 
nozzle, 242-243 
nozzle-bucket, 260-261 
stage, 259-263 
thermal, 313-314, 332 
turbine, 261-263, 313 
volumetric, 193-196 
Electricity, 2, 12 
as work, 2 


Electrolux refrigerator, 374-375 
Emissivity, 502, 505, 509 
Emissivity factor, 503-505 
Energy, chemical, 7 

internal (see Internal energy) 
kinetic, 4 
molecular, 5, 13 
potential, 4 
.stored, 3-7, 13, 44 
thermal, 6 
transitory, 3 

unavailable, change of, 109 
Energy equation, of closed system, 22-23 
of open system, 46-47 
of steady flow, 47-48 
Engine, adiabatic, 317 
reversible, 307-309 
Carnot (see Carnot engine) 
heat (see Heat engine) 
internal-combustion, 269-285 
irreversibilities in, 397 
reciprocating steam, 305-326 
vs. turbine, 322-323 
vapor, 363 

Engine efficiency, 312-313 
Engine governing, 321-322 
Enthalpy, 29, 55 

calculation of, from tables, 128-137 
changes of, for perfect gas, 175-176, 
436 

and constant-pressure heat of combus¬ 
tion, 472 
of fusion, 118 

general equations for, 160-161 
isentropic drop of, 102, 230, 313 
of sublimation, 120, 136 
of vaporization, 118 
Enthalpy-entropy chart, 101-102 

compared with Pv and Ts diagrams, 
102 

slope of lines of constant pressure on, 
101 

and steady-flow process, 102 
Entropy, 88-113 
absolute value of, 94 
calculation of, from tables, 128-137 
changes of, during irreversible proc¬ 
esses, 91-92, 97-98 
for perfect-gas system, 186-187, 436 
during reversible processes, 88-92, 
97-98 


INDEX 


549 


Entropy, as criterion of stability, 106 
extensive character of, 93 
general equations for, 160-163 
method of evaluating change of, 93 
principle of increase of, 106 
unit of, 93 

Equation of state, 12, 16 
general, 442-450 
of ideal gas, 17 
of perfect gas, 171 
primary, 150 
secondary, 160-163, 454 
Equations, general thermodynamic, of 
pure substance, 150-164 
Equilibrium, 2, 67, 106, 150, 247-248, 
444, 460, 478n. 
chemical, 478 
metastable, 247 
unstable, 247-248 
Equilibrium constant, 478 
Ericsson cycle, 298 

Erosion of turbine blades, 259, 335, 336 
Evaluation of irreversibility, 392-395 
Evaporation, 118 

into gas-vapor mixture, 212-215 
Evaporator, 363 

Events in steam-engine cycle, 306 
Exact differential, 16 
Excess air, 293, 460, 465 
Exhaust loss, 386 
Expansion, incomplete, 311, 314 
metastable, 247-248 
multiple, 320 

multistage, 192-193, 263-264 
staged, in gas-turbine power plant, 293 
steady-flow, of gases, 196-197 
unrestrained, 33, 36, 61 
Expansion ratio, 275 
Expansion valve, 365 
Explosion gas turbine, 297 
Extensive property, 11 
External combustion, applied to gas 
cycle, 297-301 
vs. internal combustion, 334 
External and internal reversibility, 67 
Extraction of steam, 342, 346 

F 

Fahrenheit temperature scale, 10 
Fanno line, 422-426, 441 
and pressure shock, 427-428 


Fans, 188, 191-192 
Film coefficient, 509-511 
Film condensation, 520 
First Law of thermodynamics, 21 

applied to chemical reactions, 458, 460, 
477 

and closed system, 21-43 
inadequacy of, 78 
and open system, 44-59 
Flame temperature, 480-482 
Flow, across boundary, 45 

effect of Reynolds number on, 415-419 
with friction, 408-432 
laminar, 409 
maximum, 236, 239, 423 
metastable, 247-249 
in pipes and ducts, 408-432 
reversible, 68, 230-241 
steady, 47-49 
streamline, 409 
in turbine blading, 238 
turbulent, 410 
velocity distribution in, 416 
viscous, 409 
Flow process, 44 
Flow steam, 308 
Flow work, 45-46 
Flue-gas analysis, 468 
Fluid, perfect, 51 
Force on turbine bucket, 252 
Forced convection, 491, 515-518 
Free air, 195 

Freedom of molecule, degrees of, 178- 
179 

Freon series of refrigerants, 369, 370 
Freon tables, 544 
Friction, 24 

coefficient of, 414, 417, 418 
effect of, 36, 241-245, 408 
increase of pressure due to, 423 
mechanical, in steam engine, 323 
and paddle-wheel process, 35 
Froude number, 430 
Fuel injection, 278, 281-282 
Fuels, 458 

constant-pressure heat of combustion 
of, 473 

ultimate analysis of, 466 
Function, psi, 110-111 
Furnace, heat transfer in, 508-509 
Fusion, 118 


550 


BASIC ENGINEERING THERMODYNAMICS 


G 

Gage pressure, 8-9 
Gas, 118 
ideal, 17, 124 
perfect, 167-197, 433-442 
“permanent,” 167 
Gas calorimeter, 472 
Gas constant, 171-173 
relation of, to specific heats, 176, 435 
universal, 172 
Gas tables, 439-442 
Gas-turbine power plant, 286, 291-296 
closed cycle for, 300-301 
Gases, approximate data for, 173 
continuous compression of (see Com¬ 
pression of gases) 
liquefaction of, 376-379 
Prandtl number of, 515 
radiant-heat transmission of, 508-509 
real, 433-457 
as refrigerants, 362 
steady-flow expansion of, 196-197 
General Electric Company, 223 n., 350 
Generalized Z chart, 451-453 
Generator, 372 

Gibbs free-energy function (zeta prop¬ 
erty), 111-113, 152, 482-483 
Gibbs-Dalton law, 203, 204, 213 
Governing, cutoff, 258, 321 
diesel-engine, 278 
throttle, 258, 321-322, 400 
turbine, 258 

Grashof number, 519-520 
Gravity, 11, 12 

H 

Haslam and Hottel, 509n. 

Head, total, 51 
Heat, 2 

of combustion, 459, 470-476 
effect of phases on, 471-472 
of formation, 476-477 
latent, 118 

mechanical equivalent of, 22 
radiation of, 488-490, 501-509 
of reaction, 6, 460 
unavailability of, 108 
unit of, 17, 21-22 
and work, 2-3 


Heat engine, 37 
Carnot, 69-72, 268 
closed-system, 37 -38 
efficiency of, 38, 332 
exhaust loss of, 386 
ideal vapor for, 349-350 
improvement in performance of, 386 
steady-flow, 53-54 
Stirling, 73-74, 268 

Heat exchange, by conduction, 488-489, 
493-501 

by convection, 488, 490-493, 509-521 
by mixing, 487 

by radiation, 488-490, 501-509 
Heat exchanger, counterflow, 524 
economics of, 487 
heat transfer in, 54 
irreversibility in, 398-399, 487 
parallel-flow, 522 
Heat flow, reversibility in, 61-65 
Heat pump, 71, 358, 379-382 
Carnot, 71 
efficiency of, 380 

improvement in performance of, 386 
Heat transfer in furnace, 508-509 
Heat transmission, 487-529 
Heater pressures, selection of, 346-347 
Heating of buildings, 379-381, 488 
Heating values, higher and lower, 474 
Helium, 173 
liquefaction of, 379 

Helmholtz free-energy function (Psi func¬ 
tion), 109 

Hershey, R. L., 480 
Hirn’s analysis, 315-319 
Holland, C. K., 247n., 248 
Horsepow r er, 36 
Horsepower-hour, 36 
Hot body, 71, 358 
Hottel, H. C., 480, 504 n., 509 n. 
Humidification, 217, 221-222 
Humidity, absolute, 227 
relative, 215-217 
specific, 215-216 
Hydraulic radius, 420 
Hydrogen, 173 
as fuel, 458 
liquefaction of, 379 

as used in Electrolux refrigerator, 
375 


INDEX 


551 


I 

Ice, 117-127 
Ideal gas, 17, 19 
Identical states, 15 
Impulse staging, 250-255 
Impulse turbine, 250 
Impulse-turbine staging, 249-254 
Incomplete expansion, 311, 314, 323 
Incompressible fluid, 51, 231 
Independent properties, 12, 16 
Indicator, 25 

Indicator diagram, 25-26, 307-311, 315 
ideal, 324 
Inhibitor, 483 
Initial condensation, 314 
Intercooling, 192, 292 
Internal combustion, 268 
vs. external combustion, 334 
Internal-combustion engine, 269-285 
effect of dissociation in, 478 
Internal-combustion turbine, 285-297 
Internal energy, 13 

calculation of, from tables, 128-137 
changes of, for perfect gas, 175-176, 
436 

and constant-volume heat of combus¬ 
tion, 471 

general equations for, 160-162 
International critical tables, 445n., 495n. 
International scale of temperature, 10 
Irreversibility, evaluation of, 386-407 
internal and external, 67, 106 
sources of, in reciprocating engine, 314- 
315 

Irreversible process, 68 

change of entropy for, 90-93 
and perfect gas, 185-186 
Isentropic process, 97, 102, 311 
for perfect gas, 181-184 
for perfect-gas mixtures, 211 
in reversible-adiabatic nozzle, 230-231 
Isothermal compression, 190-191 
Isothermal process, 31, 111, 159 
for perfect gas, 181-184 
as reversible process, 63-64, 96-97 

J 

Jet propulsion, 295-297 
Joule coefficient, 174 


Joule-Thomson coefficient, 159, 174, 377 
Joule’s law, 174 

K 

Kay, W. C., 451, 453 
Kaye, J., 439 

Keenan, J. H., 128, 129, 131-133, 136, 
138, 140, 141, 152, 410n., 439, 450n., 
53 In., 533n., 535n., 537n. 

Kelvin, 83 

Kelvin scale of temperature, 11, 85 
Keyes, F. G., 128, 129, 131-133, 136, 138, 
140, 141, 152, 410n., 450n., 531n., 
533n., 535n., 537 n. 

Kilowatt, 36 
Kilowatt-hour, 36 
Kinetic energy, 4, 44-55 
Kinetic theory of matter, 3, 169 
King, W. J., 495n. 

Kirchhoff’s law, 503 

L 

Laminar flow, 409, 415 

film coefficients for, 518-521 
Latent heat, 118 

Law of corresponding states, 450-451 
Laws of motion, 44 
Lead storage battery, 484 
Leakage, 258, 323 
of heat, 359 
Lenoir cycle, 285 
Lenoir engine, 284-285 
Linde process, 377-379 
Liquefaction of gases, 376-379 
Liquid, compressed, 128, 133-135 
incompressible, 231 
saturated, 119, 129 
Liquids, viscosity of, 411 
Load ratio, 280 

Logarithmic mean temperature differ. 

ence (LMTD), 524 
Loss, rotation, 260-261 
Losses of engine and turbine, 322-323 

M 

McAdams, W. H., 495n. 

Mach number, 414 
Mangelsdorf, 509w. 


552 


BASIC ENGINEERING THERMODYNAMICS 


Marks, L. S., 434n., 538n., 539n., 540n., 
541n., 542n., 543n. 

Mass vs. weight, In., 412 
Mass balance in combustion, 462 
Maximum flow, 236, 239, 423 
Maximum work, and availability, 388 
on piston, 25 

and psi and zeta properties, 110-113 
and reversibility, 61, 80, 388 
on turbine bucket, 252, 256 
Maximum-work process, 33 
Maxwell relations, 150-155 
Mean effective pressure, 276 
Mean temperature difference, 521-525 
arithmetical and logarithmic, 524 
Measurement of pressure, 8 
Mechanical equivalent of heat, 22 
Medium, 27, 387-403 
Mercury, in binary-vapor cycle, 350-354 
steam-flow ratio, 353 
as used for pressure measurement, 8 
Mercury tables, 538 

Metallurgical limit, 268, 288, 289, 334, 
336, 349, 358 

Metastable equilibrium, 247 
Metastable expansion, 247-248 
Methyl choride, 370 

Mixtures, of ammonia and water, 370- 
371 

critical point of, 453-454 
of perfect gases, 203-211 
of reactants, 458 
of real gases, 453-454 
saturated, of a gas and a vapor, 212- 
224 

Moisture, effect of, 259, 334 
percentage of, in exhaust steam, 337- 
339 

surface, of fuel, 466 
Molar volume, 172 
Mole, 172, 204 
artificial, 466 

combustion equations in terms of, 462 
Mole fraction, 204-205 
Molecular weight, 172 
apparent, 173, 205 
Molecule, energy storage of, 169-171 
motion of, 169-171 
structure of, 4 
Mollier, R., 101 


Mollier chart, 101-102 

(See also Enthalpy-entropy chart) 
Motion, 4, 12, 44 
Multiple expansion, 320 
Multistage expansion, 192-193, 263-264 
Multistage turbine, 249-250 
Multivapor refrigeration, 379 

N 

National Bureau of Standards, 445n., 473 
Natural convection, 490, 518-521 
Negative temperature, impossibility of, 
85 

Newton’s laws of motion, 44 
Nikuradse equation, 417n. 

Nitrogen, 173 
Nonflow process, 44 

energy equation for, 22-23 
irreversibility of, 390-395 
and perfect gas, 179-187 
Nozzle, coefficients of, 242-244 
converging, 232 
converging-diverging, 236 
critical-pressure ratio of, 235, 239-240 
efficiency of, 242-243 
exit area of, 236, 240 
expanding, 236 
formed from blading, 238 
increase of kinetic energy in, 55 
liquid, 231-233 

metastable expansion in, 247-248 
as part of turbine, 229 
perfect-gas, 233-239 
pressure shock in, 246 
real, 241-245 

reversible-adiabatic, 230-231 
throat of, 234-235 
vapor, 239-241 

velocity of sound in, 237-239, 246 
Nozzle angle, 251, 253, 256 
Nozzle-bucket efficiency, 260-261 
Nozzle condition line, 240-241 
Nucleus of condensation, 247 
Nusselt equation, 520 
Nusselt number, 514-520 

O 

Octane number, 275 
Open system, 1, 3 

energy equation of, 47 


INDEX 


553 


Open system, and First Law, 44-56 
stored energy of, 44-45 
work and, 45 
Open-system cycle, 54 
Orsat apparatus, 468 
Otto-cycle engine, 270-277 
simplified analysis of, 480 
Over-all heat-transfer coefficient, 511- 
513 

Oxygen, 173 

P 

Paddle-wheel process, 14, 35 
Parallel flow, 522 

Partial pressure, 171, 204, 374-375 
and equilibrium constant, 479 
Path, state, 13-14 
Perfect gas, 167-197 

change of, enthalpy of, 176 

entropy of, 181, 183, 184, 186-187 
internal energy of, 174-176 
continuous compression of, 188-195 
equation of state of, 171 
and Fanno line, 425 
gas constant of, 171-172 
and polytropic process, 179-184 
steady flow of, 187-188, 233-238 
Perfect gases, mixtures of, 203-211 
Perpetual-motion machine, 80 
Phase diagram, 124-125 
Phases, effect of, on heat of combustion, 
471-472 
of ice, 118 
of matter, 3 

of pure substance, 117-120 
Phi property, 439-441 
Pipe, flow through, 408-432 
Piston air compressor, 188-196 
Piston displacement, 194-195, 275, 276, 
310, 317 

Planck, Max, 78 
Point function, 156n. 

Poise, 410 

Poly tropic process, 179-184, 437 
Potential energy, 4 
Power, 36 

from combustion, 482-484 
from gas-system cycles, 268-304 
from vapor-system cycles, 327-357 
Prandtl number, 514-520 


Pressure, 8 
absolute, 9 
atmospheric, 8 
critical, 119 
gage, 9 

mean-effective, 276 
measurement of, 8 
nomenclature of, 8-9 
partial (see Partial pressure) 
reduced, 450-453 
relative, 439-440 
source of, 170 

Pressure drop in pipes, 420-422 
Pressure ratio in axial-flow compression, 
264 

Pressure shock, 246, 426 
Pressure-volume diagram, 13-14 
maximum work on, 25-26 
of water, 125 

Prime mover, 54-55, 328, 330, 331 
engine as, 55-56, 305 
losses in, 322-323 
turbine as, 54-55, 229-230 
Principle of increase of entropy, 106 
Process, 1, 14-15 
adiabatic (see Adiabatic process) 
adiabatic mixing, 207-211 
closed-system, 22 
constant-internal-energy, 32 
constant-pressure, 29, 64-65, 97-99, 
184, 436 

constant-volume, 27, 65-67, 97-99, 
184, 436 

engineering, for atmospheric air, 221- 
224 
flow, 44 

isentropic (see Isentropic process) 
isothermal (see Isothermal process) 
maximum-work, 33 
nonflow (see Nonflow process) 
paddle-wheel, 14, 35 
polytropic, 179-184, 437 
reversible (see Reversible processes) 
throttling, 143-145 
Products, 6, 458 
Propane, 370 

Properties, of mixture, 205-207 
as point functions, 156n. 
primary, 150 

of real-gas mixture, 453-454 


554 


BASIC ENGINEERING THERMODYNAMICS 


Properties, of saturated gas-vapor mix¬ 
ture, 214 

tables of, 128-137, 530-544 
Property, 7 
composite, 16 
extensive, 11 
identification of, 15 
intensive, 11 
phi, 439-441 
viscosity as, 411 
zeta, 111-113, 152, 482-483 
Proportion by volume and weight, 205 
Proportionality factor, 22 
Pseudocritical temperature and pressure, 
453-454 

Psi function, 109 
Psychrometric chart, 218-221 
lines of relative humidity on, 219 
Psychrometry, 215 
Pump work, 56, 332 

Pumps, 54, 56, 330, 332, 341, 343, 346, 
372, 374-375 
Pure substance, 117-145 
definition of, 12 

general thermodynamic equations of, 
150-164 

Q 

Quality, experimental determination of, 
142-145 

of saturated vapor, 120, 130-131 
Quasi-steady flow, 55 

R 

Radiation of heat, 488-490, 501-509 
Rankine cycle, 330-336 
Rankine temperature scale, 11, 85 
Ratio, compression, 275 
critical-pressure, 235, 239-240 
expansion, 275 
load, 280 

of mercury to steam flow in binary- 
vapor cycle, 353 

pressure, in axial-flow compression, 264 
of specific heats, 156, 173, 176-179 
work, 73, 75, 329, 331, 362, 364 
Rayleigh line, 426-430, 441 
Reactants, 6, 458 
Reaction, chemical, 6 
Reaction staging, 255-257 


Reaction turbine, 249-250, 255-259 
Real gases, 433-457 
mixtures of, 453-454 
Receiver, 192 

Reciprocating engine, 55-56, 305-326 
cylinder condensation in, 315, 319-321, 
323 

sources of irreversibility in, 314-315 
steady-flow analysis of, 55-56 
turbine compared with, 322-323 
Rectifier, 373 

Reduced pressure, temperature, and vol¬ 
ume, 450-453 
Reevaporation, 315 
Refrigerants, 368-370 
gases and vapors as, 362 
Refrigerating capacity, 359 
Refrigerating efficiency, 362 
Refrigerating machine, 71, 358-376 
improvement in performance of, 386 
Refrigeration, 358-385 
adsorption, 375-376 
ammonia-absorption, 370-375, 383-384 
ammonia-compression, 367-368 
coefficients of performance in, 360-361, 
374 

comparison of media for, 362, 368-370 
dry-compression, 365 
low-temperature, 379 
multivapor, 379 
ton of, 359 

vapor-compression, 364-368 
wet-compression, 363 
Refrigerator of heat engine, 37, 71, 74, 81, 
82, 298 

Regeneration, 67 

in ammonia-absorption refrigeration, 
373 

in gas-turbine power plant, 291-294 
in liquefaction of gases, 376-379 
in Stirling cycle, 74 
in Stirling and Ericsson cycles, 298-300 
in vapor power plant, 340-344 
Regenerative cooling, 373, 377 
Regenerative vapor cycle, 339-349 
Reheat-regenerative vapor cycle, 347, 356 
Reheat vapor cycle, 336-339 
Reheater, 337, 338 
Relative humidity, 215-217 
lines of, on psychrometric chart, 219 
Relative pressure and volume, 439-440 


INDEX 


555 


Release, 306 

Residual velocity, 253-255, 323 
Resistance, 496 

Reversibility, and availability, 388 
and closed-system process, 60-61 
external and internal, 67 
in heat flow, 61-65 
requirements for, 60-61 
in steady flow, 68 

and temperature-entropy diagram, 95- 
99 

thermodynamic, 60 
Reversible adiabatic engine, 307-309 
Reversible combustion, 483 
Reversible cycle, 69-75, 104-105, 273- 
274, 307-309 

Reversible processes, 60-68, 80, 95-99, 
388 

work and heat flow for, 179-185, 436- 
442 

Reynolds number, 414 

calculation of, using hydraulic radius, 
420 

and coefficient of friction, 414 
and convective heat transfer, 514-518 
effect of, on flow characteristics, 415- 
419 

vs. friction coefficient, 418 
Rocket, 296-297 

Rotary air compressor, 188, 191-192, 
263-264 

Rotation loss, 260-261, 323 
Roughness factor, 417n. 

S 

Saturated atmospheric air, 213-214 
Saturated liquid, 119, 129 
Saturated mixture, 213 
Saturated vapor, 119, 129 
as refrigerant, 362 
Saturation, adiabatic, 217 
Scales of temperature, 10-11 
Second Law of thermodynamics, 78-86 
and Carnot engine, 81 
Clausius’ statement of, 78 
Planck’s statement of, 78 
reversibility and, 80 
Separating calorimeter, 142 
Shaft work, 47, 51, 55, 56 
Silica gel adsorption refrigeration, 375- 
376 


Sink, 37 

Skin-friction drag, 430 
Skin temperature, 482, 511 
Solar radiation, 505, 507 
Solid, properties of, 135-137 
Solutions of ammonia in water, 370-371 
Source, 37 
of pressure, 170 

Source temperature, effective, 268, 334 
Specific heat, at constant pressure, 30, 
159, 173, 175, 434 
at constant volume, 28, 173, 175 
general thermodynamic equations for, 
155 

molar, 434, 435 
polytropic, 180 
of superheated steam, 133 
variable, and perfect gas, 433-442 
variation of, with temperature, 434 
of water, 118 

Specific heats, 155, 173, 206 
of mixture, 206 
ratio of, 156, 173, 176-179 
relation between, of perfect gas, 176, 
435 

Specific humidity, 215-216 
Specific volume, 11 
Stable equilibrium, 106 
Stage, turbine, 249-259 

effectiveness vs. efficiency of, 400-403 
irreversibility of, 402 
velocity-compound, 253-254 
Stage effectiveness, 401-403 
Stage efficiency, 259-261, 400-403 
Staging, of axial-flow compressor, 263- 
264 

comparison of impulse and reaction, 
258-259 

impulse, 250-255 
two-row, 253-255 
reaction, 255-257 
turbine, 249-257 
Standard atmosphere, 8 
State, change of, 8 
equation of, 11-12 
most stable, 106, 478 
of system, 7 
State path, 13 

Steady flow, equation of, continuity, 52- 
53 

energy, 47-48 


556 


BASIC ENGINEERING THERMODYNAMICS 


Steady flow, of fluids, 229-264 
irreversibility in, 395-400 
in pipes and ducts, 408-432 
requirements for, 47 
reversibility in, 68 

Steady-flow analysis of reciprocating 
engine, 55-56, 311 
Steady-flow heat engine, 53-55 
Steady state, 489 

Steam, in binary-vapor cycle, 350-354 
determination of moisture in, 142-145 
as refrigerant, 370 
saturated, 119, 129, 213 
superheated, 131-133, 216, 320 
supersaturated, 247-248, 444 
Steam engine, 305-326 
Steam tables, 530-537 
Stefan-Boltzmann equation, 503 
Stefan’s constant, 503 
Stirling cycle, 73, 298 
Stirling engine, 73-74, 268, 298-300 
Storage battery, 484 
Streamline flow, 409 
Su, Gong-Jen, 452 
Sublimation, 120, 137 
enthalpy of, 120, 136 
Sulphur as fuel, 458 
Sulphur dioxide, 370, 375-376 
as combustion product, 458, 461 
Sulphur dioxide tables, 541-542 
Superheated liquid, 444 
Superheated vapor, 118, 131-133 

effect of use of, on Rankine cycle, 335- 
336 

Superheater, 335, 337, 338 
Supersaturated steam, 247-248, 444 
Supersaturation limit, 248 
Surface, Gibbs, 121n. 
pvT, 121-124 
separation, 212-213 
wetted, 410 

Surface coefficient, 510n. 

Surface moisture, 466 
Surface tension, 7, 247-248 
Surroundings, 27, 358-359 
Sweigert, R. L., 178n., 434 
Systems, boundaries of, 1 
closed, 1 
conservative, 49 
fuel-injection, 278 
open, 1 


Systems, simple, 12 
thermodynamic, 1 

T 

Tables, ammonia, 539-540 
carbon dioxide, 543 
Freon, 544 
mercury, 538 
of properties, 128-137 * 
steam, 530-537 
sulphur dioxide, 541-542 
vapor, 128 

Temperature, absolute, 11, 83-85 
absolute zero of, 10-11, 85-86 
base, in combustion, 459, 472 
critical, 119, 121, 127 
definition of, 9 
effective source, 268, 334 
equality of, 9, 83 
flame, 480-482 
international scale of, 10 
maximum, as result of combustion, 460, 
461 

mean difference of, 521-525 
measurement of, 9-10, 83-85 
negative, 85 

pseudocritical pressure and, 453-454 
reduced, 450-453 
scales of, 10-11 
skin, 482 

thermodynamic scale of, 83-86 
wet- and dry-bulb, 215, 217 
Temperature-entropy diagram, and com¬ 
parison of cycles, 104, 283-284 
and irreversible process, 99-100 
regeneration shown on, 104 
and reversible cycle, 104-105 
and reversible process, 95-98 
of water, 137-140 
Temperature gradient, 491 
Temperature scales, 10-11, 83-85 
Theoretical air for combustion, 465 
Thermal efficiency, 313-314, 332 
Thermodynamic medium, 27 
Thermodynamics, 1 
of combustion, 458-486 
First Law of, 21 
Second Law of, 78-86 
Third Law of, 94 
Thermometer, 9-10, 83 
dry- and wet-bulb, 217 


INDEX 


557 


Third Law of thermodynamics, 94 
Throat, of nozzle, 234-237 
of venturi, 232 

Throttling in refrigeration, 365, 400 
Throttling calorimeter, 142-145 
Throttling governor, 258, 321-322, 400 
Throttling process, 143-145 
Ton of refrigeration, 359 
Total head, 51 

Transition from laminar to turbulent 
flow, 415-418 

Transmission of heat, 487-529 
Triple point, 120, 126-127, 369 
Turbine, 229-263 

condition curve for, 261-262 
effectiveness of, 402-403 
efficiency of (see Turbine efficiency) 
gas, 285-297 
explosion, 297 
impulse, 249 
multistage, 249-250 
reaction, 249, 255-257 
vs. reciprocating engine, 322-323 
steam, 230 

thrust on shaft of, 259 
Turbine blades, erosion of, 259, 335, 336 
Turbine blading, 238, 249-257 
Turbine bucket, 230, 249-261 
Turbine efficiency, 261-263, 313 

effect of, on moisture in exhaust, 337- 
339 

Turbine governing, 258, 321-322 
Turbine nozzle, 229-248 
Turbine staging, 249-257 
(See also Stage, turbine) 

Turbulence, 323 

effect of, on heat transfer, 515 
Turbulent flow, 410, 415 
Two-row impulse stage, 253-255 
Two-stage compression, 192-193 

U 

u, definition of, 13 
vs. e, 13, 45 

Unavailability of heat, 108 
Unavailable energy, change of, 109 
Uniflow engine, 307 

and cylinder condensation, 320-321 
Unit, of capacity in refrigeration, 359 
of heat, 21-22 
of work, 22 


Universal gas constant, 172 
Unrestrained expansion, 33, 36, 61 
Unstable equilibrium, 247-248 
Useful work, 27, 389-395 

V 

V-l bomb, 296 
Vacuum, measurement of, 9 
van der Waals equation, 162n., 442-447 
Vapor, and gas, 167 
ideal, for heat engine, 349-350 
properties of, 118-128 
as refrigerant, 362 
saturated, 119, 129 
superheated, 118, 131-132 
water, 118 

Vapor-absorption refrigeration, 370-375 
Vapor-compression refrigeration, 364-370 
compared with absorption refrigera¬ 
tion, 373 

Vapor generator, 330 
Vapor pressure, 215-217 
Vapor tables, 128-133, 135-137 
Vaporization, characteristics of, 118-120, 
212-214 
Velocity, 44 

approach, 242-243 
average, in pipe, 416 
bucket, 250-255 
coefficient of, 241-243 
exit, 251-255 

in flow through nozzle, 236, 237 
ideal, 242-244 
residual, 253-255, 323 
of sound, 237, 423 

Velocity-compound turbine stage, 253- 
254 

Velocity gradient, 410 
Velocity profile, 416 
Venturi, 232 

Viscosity, 49, 327n., 411, 441 
coefficient of, 410-411 
effect of, in steady flow, 229, 408 
expressed dimensionally, 412-413 
and viscous drag, 409-411 
of water and steam, 137 
Viscous drag, 409-411 
Viscous flow, 409 
Volume, 11 

calculation of, from tables of proper¬ 
ties, 130-131 


558 


BASIC ENGINEERING THERMODYNAMICS 


Volume, of compressed liquid, 133-135 
molar, 172 

proportion by, of mixture, 205 
reduced, 450-453 
relative, 439-440 
specific, in nozzle flow, 236, 237 
Volumetric analysis, of gas mixture, 205 
of products of combustion, 467-469 
Volumetric efficiency, 193-196 
Volumetric relations in combustion, 463 

W 

Water, as combustion product, 458 
compressed, 133-135 
critical point of, 119, 127-128 
properties of, 117-142 
as pure substance, 117-145 
as refrigerant, 370 
solid phases of, 118, 125 
surface tension of, 127, 247 
triple point of, 120, 126-127 
as vapor for heat engine, 350 
Water vapor, 118, 173 
absorption of radiant heat by, 508 
mixture of, with air, 212-224 
Weber, H. C., 452 
Weight, molecular, 172, 205 
proportion by, of mixture, 205 
Wet-bulb depression, 217 
Wet-bulb temperature, 215, 217 


Wet compression, 363, 369-370 
Wet steam, 120, 122, 130-131 
Wetted surface, 410 
Wiredrawing, 311, 314, 323 
Work, 2 

and closed system, 24 
constructive, 388 
external, 45 
flow, 45-46 
gross and net, 389 
maximum, and availability, 388 
on piston, 25 

and reversibility, 61, 80, 388 
on turbine bucket, 252, 256 
and open system, 45 
paddle-wheel, 24 
of prime mover, 55 
shaft, 47 
unit of, 22 
useful, 27, 389-395 
Work ratio, 73, 75, 329, 331, 362, 364 

Y 

Yellot, J. I., 247, 248 

Z 

Zero, absolute, of entropy, 94-95 
of temperature, 10-11, 85-86 
Zeta property, 111-113, 152, 482-483 














































